Physics High School

## Answers

**Answer 1**

The child made around 3.3 **revolutions** around the merry-go-round while he was pushing it.

We know that,

**Angular acceleration** (α) = Constant

Time (t) = 34.0 s

Final** velocity** (ω) = 1.20 rad/s

Initial velocity (ω0) = 0

First, we need to find angular displacement using the formula of angular acceleration.

α = (ω - ω0)/t

α = (1.20 - 0)/34.0

α = 0.03529411764 rad/s²

Now, let's calculate **angular displacement** (θ) using the formula of angular acceleration.

θ = ω0t + 1/2 αt²

θ = 0 + 1/2 × 0.03529411764 × (34.0)²

θ = 20.76746491 rad

As we know that 2π radians make one complete revolution.

So, the total number of revolutions made by the child can be given as,

Number of revolutions

(n) = θ / 2π

n = 20.76746491 / 2π

n = 3.30792287 revolutions (approx)

Therefore, the child made around 3.3 revolutions around the merry-go-round while he was pushing it.

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## Related Questions

Impulse: A very small 51-g steel ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for 0.50 ms. The ball rebounds elastically and returns to its original height. The total time interval for a round trip is 3.00 s. What is the magnitude of the average force exerted on the ball by the plate during contact with the plate

### Answers

The **magnitude** of the average **force** exerted on the ball by the plate during contact is approximately 0.204 N (rounded to three significant digits).

To find the magnitude of the average force exerted on the ball by the plate during **contact**, use the concept of **impulse**. Impulse is defined as the change in **momentum** of an object and can be calculated as the product of the average force and the time interval over which the force acts.

Given:

Mass of the steel ball (m) = 51 g = 0.051 kg

Contact time (Δt) = 0.50 ms = 0.50 x 10^-3 s

Total **time** for a round trip (T) = 3.00 s

First, find the **initial** and **final** **velocities** of the ball. Since the ball falls and rebounds elastically, the change in velocity during contact will be twice the velocity just before contact.

The total distance traveled during a round trip is twice the height of the ball:

2h = g * (T/2)^2

where g is the acceleration due to gravity.

The initial velocity (v_initial) of the ball just before contact can be found using the formula:

v_initial = sqrt(2gh)

where h is the height of the ball.

The final velocity (v_final) of the ball just after rebounding is equal in magnitude but opposite in direction to the initial velocity.

The change in velocity (Δv) during contact is given by:

Δv = 2 * v_initial

Now, calculate the average force (F_avg) exerted on the ball using the impulse formula:

Impulse = F_avg * Δt = Δp

where Δp is the change in momentum.

The change in momentum (Δp) is equal to the mass of the ball times the change in velocity:

Δp = m * Δv

Putting it all together:

F_avg * Δt = m * Δv

F_avg = (m * Δv) / Δt

Now, substitute the given values and calculate the magnitude of the average force exerted on the ball by the plate during contact.

The magnitude of the average force exerted on the ball by the plate during contact is approximately 0.204 N (rounded to three significant digits).

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g What is the magnitude of a the vertical electric field that will balance the weight of a plastic sphere of mass 2.1 g that has been charged to -3.0 nC? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)

### Answers

the **magnitude** of the vertical electric field that will balance the **weight** of a plastic sphere of mass 2.1 g that has been charged to -3.0 nC is 2.26 x 10^3 N/C.\that has been charged to -3.0 nC is 2.26 x 10^3 N/C.Given values:

Mass of plastic sphere, m = 2.1 g = 0.0021 kg

Charge on plastic sphere, q = -3.0 nC = -3.0 x 10^-9 CThe force of gravity or weight acting on the sphere is given by the formula;Fg = mgwhere,

Fg = force due to gravity, m = mass of the object, g = acceleration due to gravityIf Fg is equal to the force due to the electric field, Fe, the plastic sphere will be in **equilibrium**.

Hence, the electric field needed to balance the weight of the plastic sphere can be calculated using the formula;Fe = Fg = mgHence,

we can write the following;Fe = mg= (0.0021 kg)(9.81 m/s²)= 2.0601 x 10^-2 NFe = (k*q)/(r²)

where k = Coulomb’s constant and r is the radius of the sphereFe = (9.0 × 10^9 N ∙ m²/C²)(-3.0 × 10^-9 C)/(r²)

To solve for the radius of the plastic sphere, we must first find its **volume** which is given by

;V = (4/3)πr³V = (4/3)π(0.001 r)³ = (4/3)π(1.0 × 10^-9 m³) = (4 × 3.14 × 10^-9 m³)/3 = 4.186 x 10^-9 m³

Now, we can calculate the radius of the **sphere** using the formula;Density of the plastic sphere,

ρ = m/V = (0.0021 kg)/(4.186 x 10^-9 m³) = 5.016 x 10^5 kg/m³

We can then find the radius, r of the sphere using the formula;m = ρV = (5.016 x 10^5 kg/m³)(4/3)πr³ = 2.105 x 10^-3 kg;

(given that the mass of the sphere, m = 2.1 g) => r = 4.05 x 10^-5 m

Substitute the value of r into the equation for the electric field to get the following;

Fe = (9.0 × 10^9 N ∙ m²/C²)(-3.0 × 10^-9 C)/(4.05 x 10^-5 m)² = 2.26 x 10^3 N/C

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A Ferris wheel is a vertical, circular amusem*nt ride with radius 6 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 6 s. Consider a rider whose mass is 50 kg.

At the bottom of the ride, what is the rate of change of the rider's momentum? (dp^^->)/(dt) = kg Â· m/s/s

At the bottom of the ride, what is the vector gravitational force exerted by the Earth on the rider? vector Fgrav = N

At the bottom of the ride, what is the vector force exerted by the seat on the rider? vector Fby seat = N

Next consider the situation at the top of the ride. At the top of the ride, what is the rate of change of the rider's momentum? (dp^^->)/(dt) = kg Â· m/s/s

At the top of the ride, what is the vector gravitational force exerted by the Earth on the rider? vector Fgrav = N

At the top of the ride, what is the vector force exerted by the seat on the rider? vector Fby seat = N

A rider feels heavier if the electric, interatomic contact force of the seat on the rider is larger than the rider's weight mg (and the rider sinks more deeply into the seat cushion). A rider feels lighter if the contact force of the seat is smaller than the rider's weight (and the rider does not sink as far into the seat cushion).

Does a rider feel heavier or lighter at the bottom of a Ferris wheel ride? lighter heavier Does a rider feel heavier or lighter at the top of a Ferris wheel ride? lighter heavier

### Answers

A Ferris wheel is a vertical,** circular amusem*nt** ride with radius 6 m. Riders sit on seats that swivel to remain horizontal. The **Ferris** wheel rotates at a constant rate, going around once in 6 s. Consider a rider whose mass is 50 kg.

At the bottom of the ride, the rate of change of the rider's momentum is (dp/dt) = 0 kg·m/s^2 because the direction of the rider's velocity is not changing.At the bottom of the ride, the vector **gravitational** force exerted by the Earth on the rider is Fgrav = mg = (50 kg)(9.8 m/s^2) = 490 N downward.At the bottom of the ride, the vector force exerted by the seat on the rider is Fby seat = (mg + ma) upward, where a is the acceleration due to the normal force. The normal force of the seat is the force exerted by the seat on the rider and is equal in magnitude to the force exerted by the rider on the seat (by **Newton's third law**).

Therefore, the rider feels heavier if the normal force is greater than the weight, and lighter if the **normal** force is less than the weight.

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The starter motor of a car engine draws an electric current of 22.0 A from the battery. The copper wire to the motor is 5.20 mm in diameter and 1.00 m long. The starter motor runs for 0.75 s before the car engine starts up. How much electric charge passes through the starter motor?

### Answers

**Answer:**

Q:Question:

The starter motor of a car engine draws an electric current of 20.0 A from the battery. The copper wire to the motor is 4.30 mm in diameter and 1.49 m long. The starter motor runs for 0.75 s before the car engine starts up. How much electric charge passes through the starter motor? What is the current density in the wire?

A:Answer:

a car on a wet road can achieve an acceleration of only -1.00m/s^2 without sliding. Find the required stopping distance for a speed of 48km/hr

### Answers

The car on a wet road with an **acceleration** of -1.00m/s^2 without sliding requires a stopping distance of 17.78 meters for a speed of 48km/hr.

In order to calculate the required stopping distance, we need to first convert the **speed** of 48km/hr to m/s. 48km/hr is equivalent to 13.33m/s.

Next, we can use the formula d = (v^2)/(2a) to find the stopping **distance**.

Plugging in the values we get, d = (13.33^2)/(2*-1.00) = 17.78 **meters**.

Therefore, for a car on a **wet** road with an acceleration of -1.00m/s^2 without sliding, it will require a distance of 17.78 meters to come to a complete stop from a speed of 48km/hr.

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design a quarter-wave matching transformer to match a 10 ω load to a 75 ω system. assume a design frequency of 200 mhz.

### Answers

The **electrical length** of the quarter-wave transformer is λ/4 = 0.375 meters.

To design a quarter-wave matching transformer to match a 10 Ω load to a 75 Ω system at a design frequency of 200 MHz, the following steps can be followed.

First, calculate the characteristic impedance of the transmission line using the formula: Z0 = sqrt(ZL * ZS), where ZL is the load **impedance **(10 Ω) and ZS is the system impedance (75 Ω). Thus, Z0 = sqrt(10 Ω * 75 Ω) = 27.39 Ω.

Next, determine the electrical length of the quarter-wave transformer.

For a quarter-wave at 200 MHz, the **wavelength **is λ = c / f = (3e8 m/s) / (200e6 Hz) = 1.5 meters.

Finally, construct a transmission line with a characteristic impedance of 27.39 Ω and an electrical length of 0.375 meters.

This will act as the quarter-wave matching transformer, effectively matching the 10 Ω load to the 75 Ω system at the desired **frequency **of 200 MHz.

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Olivia rubs an inflated balloon against a blanket and then "sticks" the balloon to the wall (neutral) where it stays without moving. The reason the charged balloon stuck to a wall is because

### Answers

The reason the charged balloon stuck to a wall is because the **frictional force** generated during the rubbing of the balloon against the blanket led to the buildup of static electricity on the surface of the balloon.

Olivia rubs an inflated balloon against a blanket and then "sticks" the balloon to the wall (neutral) where it stays without moving. The reason the charged balloon **stuck** to a wall is because the frictional force generated during the rubbing of the balloon against the blanket led to the buildup of static electricity on the surface of the balloon. As a result, the balloon becomes electrically charged and can be used to demonstrate basic principles of electrostatics.The charged balloon adheres to the wall because of the attraction between opposite charges. In this case, the wall may have a neutral or slightly charged surface, and when the charged **balloon** is brought near it, it induces an opposite charge on the wall. This attraction between the opposite charges allows the balloon to stick to the wall until the charge dissipates or is neutralized.

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An electron is accelerated through a potential difference of 1.9 kV and directed into a region between two parallel plates separated by 20 mm with a potential difference of 170 V between them. The electron is moving perpendicular to the electric field when it enters the region between the plates. What magnetic field is necessary perpendicular to both the electron path and the electric field so that the electron travels in a straight line

### Answers

To make the **electron** travel in a straight line between the parallel plates, the magnetic field must exert a force on the electron that cancels out the force due to the electric field.

The force experienced by a charged particle moving in a magnetic field is given by the equation** F = qvB, **where F is the force, q is the charge of the particle, v is the velocity, and B is the magnetic field strength.

In order to cancel out this force, the magnetic field strength B must satisfy the condition Fe = F. Therefore, we have qE = qvB, which simplifies to** E = vB.**

Given that the electron is accelerated through a potential difference of 1.9 kV and directed into the region between the plates, we can calculate the electric field strength using the formula E = V/d, where V is the potential difference and d is the distance between the plates. Therefore, E = 170 V / 0.02 m = **8500 V/m**.

Now we can substitute this value into the equation E = vB to solve for the required magnetic field strength B. Since the electron is moving perpendicular to both the electric field and the magnetic field, the velocity v can be considered constant. Therefore, B = E/v =** 8500 V/m /v.**

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A Gaussian surface contains a single charge within it, and as a result an electric flux passes through the surface. Suppose that the charge is then moved to another spot within the Gaussian surface. Does the flux through the surface change

### Answers

A Gaussian surface contains a single **charge **within it, and as a result an electric flux passes through the surface. Suppose that the charge is then moved to another spot within the Gaussian surface.

Does the flux through the surface change?

The answer is that the flux through the surface will not change when the charge is moved to another spot within the **Gaussian surface**.

This is because electric flux is only dependent on the charge enclosed within the surface, not its location. In other words, the electric flux through a Gaussian surface is proportional to the charge enclosed within the surface and is independent of the **shape or size** of the surface or the position of the charge inside the surface.

So, as long as the charge remains enclosed within the surface, the flux through the surface will remain the same, regardless of the position of the charge inside the surface.

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When you push a piece of chalk across a chalkboard, it sometimes screeches because it rapidly alternates between slipping and sticking to the board. Describe this process in more detail, in particular, explaining how it is related to the fact that kinetic friction is less than static friction. (The same slip-grab process occurs when tires screech on pavement.)

### Answers

The screeching sound produced by chalk or tires on pavement is due to the alternating slipping and sticking caused by the difference between static friction and** kinetic friction**.

When an object is in motion, it experiences kinetic friction, while static friction acts on it when it is at **rest**. The force of static friction is greater than kinetic friction, which means that it takes more force to initiate motion than to maintain it.

When a piece of chalk or a tire is pushed across a surface, it initially experiences static friction, which causes it to stick to the surface. However, if the **force** applied is greater than the maximum static friction force, the object will start to slip.

Once this happens, the frictional force decreases to the level of kinetic friction, which is less than static **friction**.

This allows the object to slide more easily across the surface until it again encounters a **region** of the surface where static friction is acting on it, causing it to stick again. This back-and-forth motion creates the slipping and sticking effect that produces the screeching sound.

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Read the discussion of elastic potential energy on page 324. In the examples of the Achilles tendon, the percentage of energy lost when a runners foot strikes the ground is

### Answers

The stored **energy **in an object that has been stretched or deformed is known as elastic potential energy. The tendon of Achilles serves as a sturdy fibrous structure that links the heel bone to the** calf muscles. **

What is **elastic potential energy**

When a **runner's** foot hits the surface, the Achilles tendon experiences distortion since it lengthens to soak up the shock.

As the runner's foot hits the ground, a portion of the **energy **is converted into potential energy and stored in the elastic traits of the Achilles tendon. The energy that was initially stored in the tendon is subsequently unleashed when it rebounds and resumes its initial form, thereby helping the runner move ahead.

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Newton's law of cooling states that the rate of cooling an object is proportional to the temperature difference between the object and the surrounding medium (air).

a. True

b. False

### Answers

False, the **statement** is not true about the Law of cooling according to Isaac **Newton**.

What is Newton's Law of cooling?

According to Newton's law of cooling, which is not just applicable to air, the rate of **cooling** of an item is proportional to the temperature differential between the object and its **surroundings**.

Regardless of the medium, the law outlines how the temperature difference between the object and its **surroundings** affects the cooling process.

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Newton's law of cooling states that the rate of cooling an object is proportional to the **temperature** difference between the object and the surrounding medium (air).

This statement is false.

Option B is correct.

State Newton's Law of cooling?

** Newton's law of cooling states** that the rate of cooling of an item is proportional to the temperature differential between the object and its surroundings and is not just applicable to** air. **

The law describes how the **temperature** difference between the object and its surrounds impacts the cooling process regardless of the **medium.**

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A 2.00kg Ball Is Moving At 3.00m/S Toward The Right. It Collides Elastically With A 4.00kg Ball That Is Initially At Rest. Determine The Velocities Of The Balls After The Collision.

### Answers

Answer:

I got -3m/s

Explanation:

For this, we'll have to use the formula;

Total momentum before = Total momentum after

Since this is an elastic collision, the balls' velocity will be a negative value.

Substitute the values into the equation like so;

(2 × 3) + (4 × 0) = (2 × -v) + (4 × v)

6 + 0 = (-2v) + 4v

6 = -2v

6/-2 = v

∴ v = -3m/s

I hope this helps! Please let me know if I have any misconceptions or miscalculations!

When the charged rod touches a conducting object, the extra charges in the contact area physically move or flow to the conducting object.

a. True

b. False

### Answers

When a charged rod touches a **conducting object,** the extra charges in the contact area physically move or flow to the conducting object. Therefore, the statement is true.

When a charged rod comes into contact with a conducting object, such as a metal surface, the excess charges on the rod interact with the conducting material. Conductors allow for the free movement of charges within their structure. As a result, the extra charges on the rod, which are in close proximity to the conducting object, are able to physically move or flow to the conducting object. This process is known as **charge transfer** or **charge redistribution**. The conducting object acts as a pathway for the excess charges to distribute and reach a state of equilibrium, balancing the charge distribution between the rod and the object.

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What is the length of this tube if its fundamental frequency equals a typical speech frequency of 180 Hz

### Answers

The **length **of the tube can be determined by the formula: length = (speed of sound) / (4 x **frequency**). the length of the tube is approximately 0.476 meters when its fundamental frequency equals a typical speech frequency of 180 Hz.

To find the length of the **tube**, we can use the formula length = (speed of sound) / (4 x frequency). The **speed of sound** in air is approximately 343 meters per second at room **temperature**. Given that the** fundamental** **frequency **is 180 Hz, we can substitute these values into the formula.

Length = 343 m/s / (4 x 180 Hz)

Simplifying the equation:

Length = 343 m/s / 720 Hz

Length ≈ 0.476 meters

Therefore, the length of the tube is **approximately **0.476 meters when its fundamental frequency equals a typical **speech frequency** of 180 Hz.

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Determine how the electric field is different from that of the positive charge. Describe the differences in the electric field due to a negative charge as compared to a positive charge

### Answers

The difference between the** electric field **and positive chargeAn electric field refers to a region where an electric charge experiences an electric force due to the presence of other **charges**.

A positive charge in an electric field is repelled from positively charged objects and attracted to negatively charged objects. Its direction is the opposite of the electric field because it is positively charged. An electric field is also the force exerted on a unit charge, while a positive charge is an object that has excess protons in its nucleus. The difference in the electric field due to a **negative **charge compared to a positive charge

A negative charge is an object that has excess electrons in its nucleus, which creates an attractive force between it and positively charged objects. Negatively charged objects are **attracted **to positively charged objects in an electric field. The direction of the electric field is the same as the direction of the electric force that acts on a negative charge.

A negative charge in an electric field will always be attracted to positively charged objects and **repelled **from negatively charged objects, which is the opposite of a positive charge. Hence, the difference in the electric field due to a negative charge compared to a positive charge is that a negative charge is attracted to positive charges, while a positive charge is repelled by other positive charges and attracted to negative charges.

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A basketball player makes a jump shot. The 0.590-kg ball is released at a height of 1.90 m above the floor with a speed of 7.04 m/s. The ball goes through the net 3.02 m above the floor at a speed of 4.28 m/s. What is the work done on the ball by air resistance, a nonconservative force

### Answers

When a basketball player makes a **jump** shot, the ball goes through the net. The given information is as follows: Mass of the ball, m = 0.590 kgInitial height of the ball, h = 1.90 mInitial velocity of the ball, u = 7.04 m/sFinal height of the ball, h’ = 3.02 mFinal velocity of the ball, v = 4.28 m/sForce acting on the ball by air **resistance** is a nonconservative force.

According to the law of **conservation** of energy, energy can neither be created nor destroyed, but it can be transformed from one form to another. The ball has **gravitational** potential energy at the initial height and kinetic energy when it reaches the net.

The energy lost due to nonconservative **forces** such as air resistance is transformed into other forms of energy such as thermal energy. The work done by nonconservative forces is defined as the difference in mechanical energy from the beginning of the path to the end of the **path**. In other words, the work done by air resistance is equal to the difference between the mechanical energies at the beginning and at the end of the path.

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The receiver of a parabolic satellite dish is set up to be 3 inches from the vertex of the dish. The satellite dish is 36 inches wide. What is the depth of the satellite dish (in inches, no label necessary)?

### Answers

The depth of the **satellite dish** is 1 inch. Given that the receiver of a parabolic satellite dish is set up to be 3 inches from the vertex of the dish and the dish is 36 inches wide. We can calculate the depth of the satellite dish (in inches, no label necessary).

A parabolic **antenna **is an antenna that uses a parabolic reflector, a curved surface with the cross-sectional shape of a parabola, to direct the radio waves. The depth of the satellite dish can be calculated using the formula below: Depth = 4 * (Receiver Distance from Vertex)² / Dish Width Where; Dish Width = 36 inches Receiver **Distance **from Vertex = 3 inches

Therefore; Depth = 4 * (3)² / 36= 4 * 9 / 36= 36 / 36= 1 inch

Thus, the depth of the satellite dish is 1 inch.

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The receiver of a **parabolic satellite** dish is set up to be 3 inches from the vertex of the dish. The satellite dish is 36 inches wide then the depth of the satellite dish would be 0 inches.

Step-by-step explanation:

We know that the parabolic satellite dish is 36 inches wide. So, we can say that the diameter of the satellite dish is 36 inches.We also know that the receiver of the parabolic satellite dish is set up to be 3 inches from the vertex of the dish.The vertex of a parabola is halfway between the focus and the **directrix.** The focus is located on the axis of symmetry and is located at a distance of p from the vertex and the directrix is a line that is located at a distance of p from the **vertex**.So, the receiver of the parabolic satellite dish is located at a distance of 3 inches from the vertex and the width of the satellite dish is 36 inches. We can use these values to find the depth of the dish.

We know that the equation of a **parabola** that opens upward is given by

y = a(x - h)² + k,

where

(h, k) is the vertex of the parabola and "a" is a constant that determines how quickly the parabola opens up. Since the vertex of the satellite dish is at (0, 0), the equation of the parabola that represents the satellite dish is given by

y = ax².We know that the receiver is located at a distance of 3 inches from the vertex.

So, we can say that the x-coordinate of the receiver is 3. Therefore, we can say that the receiver is located at the point (3, 0).We also know that the diameter of the **satellite dish** is 36 inches. So, we can say that the width of the satellite dish is 36 inches and the x-coordinate of the edge of the dish is 18. Therefore, we can say that the edge of the dish is located at the point (18, y).To find the depth of the dish, we need to find the y-coordinate of the edge of the dish. We can use the fact that the receiver is located on the parabola to find the value of "a".We know that the receiver is located at the point (3, 0), which means that y = ax² = 0 when x = 3.

Substituting these values into the equation, we get:

0 = a(3)²0 = 9a

Dividing both sides by 9,

we get: a = 0

Therefore, the equation of the parabola that represents the satellite dish is y = 0x², which simplifies to y = 0.Since the equation of the dish is y = 0, we know that the y-coordinate of the edge of the dish is 0.

Therefore, the depth of the dish is 0 - 0 = 0 inches. Hence, the depth of the satellite dish is 0 inches.

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You lay a rectangular board on the horizontal floor and then tilt the board about one edge until it slopes at angle 0 with the horizontal. Choose your origin at one of the two corners that touch the floor, the x axis pointing along the bottom edge of the board, the y axis pointing up the slope, and the z axis normal to the board. You now kick a frictionless puck that is resting at 0 so that it slides across the board with initial velocity

v|t=0 = vo_x + vo_y

Write down Newton's second law using the given coordinates and then find how long the puck takes to return to the floor level and how far it is from 0 when it does so.

### Answers

The **distance **from 0 is d = **sqrt((v0_xt)2 + (v0_y2/2g)2).**

The **force **acting on the puck is due to **gravity**, and this force can be written as follows:

F = -mg j N, where j N is the normal vector. The -v e sign indicates that the force is acting in the negative direction of the y-axis. **Newton's second law **states that F = ma,

which is given as follows :

ma = -mg j N or a = -g j N . .. equation 1Let us assume that the distance from 0 is d at any time t. Then, the coordinates of the puck can be written as (x(t), y(t), z(t)) = (v0_xt, y(t), -gt2/2).If the puck reaches the floor level, then z(t) = 0. Therefore,-gt2/2 = 0 or t = sqrt(2z/g) . .. equation 2where z is the initial height of the puck above the floor level.

Using equation

1, we have a = -g j N = -gcosθi - gsinθj . .. equation 3where θ is the angle of inclination of the board. Let us assume that the initial velocity of the puck is v0 = v0_xi + v0_yj. Then, the velocity of the puck at any time t is v(t) = v0 - g t j . .. equation 4If the puck returns to the floor level, then y(t) = 0.

Therefore, v0_yt - gt2/2 = 0. We can rewrite this equation as follows:

t = 2v0_y/g . .. equation 5Substituting equation 5 in equation 4, we get v(t) = v0 - (2v0_y/g)g j = v0 - 2v0_yj . .. equation 6The final position of the puck when it returns to the floor level is given by (x(t), y(t), z(t)) = (v0_xt, 0, -g/2(2v0_y/g)2) = (v0_xt, 0, -v0_y2/2g).

Therefore, the distance from 0 is d = sqrt((v0_xt)2 + (v0_y2/2g)2).

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Joel and Marissa are on skates on an icy pond. They are facing each other with their hands touching. They then push away from each other with their hands. The force of Joel pushing upon Marissa is equal to _____. *

### Answers

If Marissa has less mass than Joel, Marissa will be **accelerated **at a faster rate than Joel. If Joel has less mass than Marissa, Joel will be accelerated at a faster rate than Marissa.

When Joel and Marissa push away from each other on skates on an icy pond, the force of Joel **pushing **upon Marissa is equal to the force of Marissa pushing on Joel. This is known as Newton's Third Law of Motion, which states that for every action, there is an equal and **opposite **reaction. Therefore, the force that Joel exerts on Marissa is exactly the same as the force that Marissa exerts on Joel. This is true regardless of whether the two individuals are stationary or moving in a particular direction. It's just the force applied by Joel and Marissa on each other.However, if the mass of Joel and Marissa is different, the accelerations of Joel and **Marissa **will be different. If Marissa has less mass than Joel, Marissa will be accelerated at a faster rate than Joel. If Joel has less mass than Marissa, Joel will be accelerated at a faster rate than Marissa.

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I was driving along at 20 m/sm/s, trying to change a CD and not watching where I was going. When I looked up, I found myself 44 mm from a railroad crossing. And wouldn't you know it, a train moving at 30 m/sm/s was only 60 mm from the crossing. In a split second, I realized that the train was going to beat me to the crossing and that I didn't have enough distance to stop. My only hope was to accelerate enough to cross the tracks before the train arrived.

Required:

If my reaction time before starting to accelerate was 0.49s , what minimum acceleration did my car need for me to be here today writing these words?

### Answers

Given that the **driver** is travelling at a speed of 20 m/s and the train is moving at 30 m/s, and there is a distance of 44 m between the driver and the railway crossing, he has only a few seconds to react to save his life. It is also given that the driver's **reaction** time is 0.49s.

We need to find out the minimum **acceleration** required for the driver to be able to cross the track before the train arrives. Mathematical approach :The minimum acceleration required can be calculated using the following **formula**: `s = ut + 0.5at^2`Here, `s = 44m`, `u = 20m/s`, `t = 0.49s`, and we need to find `a` .

Assuming the car's acceleration rate is **constant**, we can use the following formula to calculate the minimum acceleration required: `v = u + at`, where `v = 30m/s`, `u = 20m/s`, and `t = 0.49s`Substituting the above values in the above equation, we get: `30 = 20 + a(0.49)`On solving for `a`, we get: `a = 20.408 m/s^2`.Therefore, the minimum acceleration required for the driver to be able to cross the **track** before the train arrives is 20.408 m/s^2.

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A proton accelerates from rest in a uniform electric field of 630 N/C. At one later moment, its speed is 1.60 Mm/s (nonrelativistic because v is much less than the speed of light).

Required:

a. Find the acceleration of the proton. m/s^2

b. Over what time interval does the proton reach this speed

### Answers

The time interval during which the **proton **reaches this **speed **is 2.66 x 10^-5 s.

Given:

** Electric field** E = 630 N/C, Initial **velocity **of proton u = 0, Final velocity of proton v = 1.6 x 10^6 m/s a. Acceleration of the proton: The force on the proton is given by, F = q E Where q is the charge of proton and E is electric field F = (1.602 x 10^−19 C) (630 N/C) = 1.007 x 10^−16 N (magnitude)Thus, by **Newton’s second law of motion,** F = ma1.007 x 10^−16 N = (1.6726 x 10^−27 kg) aa = 6.01 x 10^10 m/s²

Therefore, the acceleration of the proton is 6.01 x 10^10 m/s².b. Time interval during which the proton reaches this speed :v = u + at Where v is the final velocity, u is the initial velocity, a is the acceleration and t is time interval. v = 0 + (6.01 x 10^10 m/s²) tv = 1.60 x 10^6 m t = 1.60 x 10^6 m / (6.01 x 10^10 m/s²)t = 2.66 x 10^−5 s

Therefore, the time interval during which the proton reaches this speed is 2.66 x 10^-5 s.

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If the children from Beijing did in fact have a demyelinating disease, what would have been the result of Dr. McKhann's conduction velocity test

### Answers

If the children from Beijing had a demyelinating disease, the result of Dr. McKhann's conduction velocity test would likely show a **slower conduction velocity.**

**Demyelinating diseases,** such as multiple sclerosis, involve damage to the myelin sheath that surrounds nerve fibers. The **myelin sheath **plays a crucial role in facilitating the rapid conduction of nerve impulses. When it is damaged, the conduction of electrical signals along the nerves is **slowed down.**

Dr. McKhann's conduction velocity test measures the speed at which electrical signals travel along the nerves. Normally, healthy nerves with intact myelin sheaths have faster conduction velocities. However, in the case of demyelinating diseases, the damaged myelin sheath interferes with the efficient transmission of electrical signals, resulting in a slower conduction velocity.

Therefore, if the children from Beijing had a demyelinating disease, Dr. **McKhann's conduction velocity test **would likely indicate a slower conduction velocity compared to individuals without the disease. This result would provide important evidence of the presence of a demyelinating disease in the children.

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5. An astronaut is rotated in a centrifuge of radius 5.2 m. (a) What is the speed if the acceleration is 6.8g

### Answers

**Explanation:**

v = √6.8gr

v = √ 6.8 X 9.8 X 5.2

v = √346.528

v= 18.6

v = 19

A projectile is launched at a 50 degree angle. What can I do to increase the time the projectile is in the air

### Answers

To increase the time a projectile is in the air, the launch angle should be decreased. A projectile is an object that is given an initial **velocity **and then allowed to move under the influence of gravity. The time of flight of a projectile depends on its initial velocity, the angle of launch, and the acceleration due to gravity.

It is also the time that an object takes to travel from its initial position to its final position. There are a few things that can be done to increase the time of flight of a **projectile**. One of the main things is to decrease the angle of launch. When the launch angle is decreased, the projectile is given a more horizontal **velocity**. This means that it will take longer to reach its peak height, and then will take longer to fall back down to the ground. The formula for the time of flight of a projectile is given as T = 2V₀sin(θ)/g.

Where T is the time of flight, V₀ is the initial velocity, θ is the launch angle, and g is the **acceleration **due to gravity. From this formula, we can see that the time of flight is directly proportional to the sine of the launch angle. So, to increase the time of flight, the launch angle should be **decreased**. This will give the projectile a more horizontal velocity, which will make it take longer to reach its peak height and fall back down to the ground.

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Tehra's obstetrician is using a medical procedure that employs high-frequency sound waves to detect the outline of the fetus and to observe its movements. This is called

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Tehra's obstetrician is using a medical **procedure** that employs high-frequency sound waves to detect the outline of the fetus and to observe its movements. This is called **ultrasonography**.

Ultrasonography is a diagnostic medical imaging method that employs high-frequency sound waves to produce images of organs and tissues in the body. The procedure is also known as ultrasound, and it is often used to visualize the fetus during **pregnancy**.Obstetricians utilize ultrasound to diagnose problems with the pregnancy and to **monitor** the fetus's growth and development. It is also used to confirm the due date of the baby, to monitor the amount of amniotic fluid present, and to detect multiple fetuses in the uterus.

During the ultrasound, a gel is applied to the skin over the uterus, and a small instrument called a transducer is moved across the skin. The **transducer** emits sound waves that pass through the body and are reflected back to the transducer when they come into contact with tissue. The reflected waves are then transformed into an image that is displayed on a monitor.Ultrasonography is a safe and painless procedure that does not use radiation, making it ideal for monitoring the health of a developing fetus. Tehra's **obstetrician** is using ultrasonography to observe the outline of the fetus and to monitor its movements.

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What is an eclipse and what are the similarities and differences between a solar eclipse and a lunar eclipse

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Eclipse is a natural phenomenon in which the light from a celestial body is blocked by another body. The two most common eclipses are solar and lunar eclipses. A Solar **Eclipse **happens when the Moon passes between the Sun and Earth.

The Moon blocks out the light from the Sun, resulting in darkness. The Moon casts a shadow on the Earth's surface when it passes in front of the Sun. A Lunar **Eclipse **occurs when the Earth comes between the Sun and the Moon. It causes the Earth's shadow to fall on the Moon, resulting in a darkened Moon. The Lunar Eclipse is not visible in all locations, and its appearance differs based on the location of the observer. It is possible to witness this event from any place on Earth where the Moon is visible. The similarities between a solar eclipse and a lunar eclipse are that they are both natural phenomena, and both involve the alignment of three **celestial bodies **(the Sun, Moon, and Earth). The Moon is always present during these two eclipses.

In addition, both of these events are predictable and have a pattern of occurrence. The differences between a solar eclipse and a lunar eclipse include the following: Solar Eclipse **Lunar Eclipse **A Solar Eclipse occurs when the Moon passes between the Sun and Earth. A Lunar Eclipse occurs when the Earth comes between the Sun and the Moon. The Moon's shadow falls on the Earth's surface. The Earth's shadow falls on the Moon. The duration of a Solar Eclipse is shorter than that of a Lunar Eclipse. The duration of a Lunar Eclipse is longer than that of a Solar Eclipse. A Solar Eclipse can only be seen from a limited **geographic **location. A Lunar Eclipse can be seen from a larger geographic location.

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You swing a bat and hit a heavy box with a force of 1500 N. The force the box exerts on the bat is Group of answer choices

### Answers

The **force **the box exerts on the bat is 1500 N.A force is a push or pull on an object that can cause it to change its velocity. Newtons (N) are the unit of measurement for force.

A force may cause an object to **accelerate**, decelerate, or change direction, for example, a ball dropped from a height accelerates towards the ground because of the force of **gravity **exerted on it. To get started on this problem, we'll need to know that the force applied to the heavy box by the bat is 1500 N.

The force the box exerts on the bat is 1500 N. When a force is **applied **to an object, an equal and opposite force is applied to the object **exerting **the force. If we hit a heavy box with a bat, the bat exerts a force on the box, and the box exerts an equal and opposite force on the bat according to Newton's third law of motion.

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in the "capacitors" lab, you learned that capacitors in parallel sum linearly and that capacitors in series are added as the sum of reciprocals. resistors are summed in the opposite ways. explain why.

### Answers

Capacitors in parallel sum linearly because each capacitor has its own separate path for the flow of charge, while capacitors in series are added as the sum of reciprocals because the total **capacitance **depends on the combined effect of reducing the overall charge storage capacity.

In parallel, each capacitor is connected across the same voltage source, and they have their own separate paths for the flow of charge. When capacitors are connected in **parallel**, the total capacitance is the sum of the individual capacitances. This is because each capacitor contributes to the overall charge storage capacity, resulting in a linear sum of capacitances.

In contrast, when capacitors are connected in series, they share the same charge between them. The voltage across each capacitor is the same, but the charge stored is divided among the capacitors. The total capacitance of capacitors in **series** is less than that of any individual capacitor because the combined effect reduces the overall charge storage capacity. The reciprocal relationship arises from the fact that the total inverse capacitance is equal to the sum of the reciprocals of the individual capacitances.

For resistors, the situation is the opposite. In parallel, each resistor has the same voltage drop, and the total resistance is the reciprocal sum of the individual resistances. This is because adding more resistors in parallel creates additional paths for current flow, resulting in a lower overall resistance. In series, the current passing through each** resistor** is the same, and the total resistance is the sum of the individual resistances because they impede the current flow sequentially.

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One end of a rope is tied to a block of mass 3.4 kg and you pull straight up on the other end of the rope to lift the block up off of the ground with an acceleration of 3.6 m/s2. What force (in N) does the rope exert on the block

### Answers

Rounded to two decimal places, this is equal to 37.24 N. The force that the rope exerts on the block of **mass **3.4 kg when it is pulled up off the ground with an **acceleration **of [tex]3.6 m/s²[/tex] is 37.24 N (Newton).

We can calculate the force exerted by the rope on the block by using Newton's second law, which states that the force is equal to mass times acceleration:

F = m * a

Where:

F = force (in Newtons)

m = mass (in kilograms)

a = acceleration (in meters per second squared)

So, in this case, we can plug in the given values to get:

[tex]F = 3.4 kg * 3.6 m/s²[/tex]

F = 12.24 N

This is the force required to accelerate the block at the given rate. However, this is not the force exerted by the rope on the block. To find that, we need to consider the fact that the force required to accelerate the block is equal and opposite to the force exerted by the rope on the block, according to **Newton's third law**. So, the force exerted by the rope on the block is also 12.24 N, but in the opposite direction.Now, if we want to find the force that the rope exerts on the block when it is pulled up off the ground with an acceleration of[tex]3.6 m/s²[/tex], we need to account for the weight of the block as well. When the block is lifted, it experiences two forces: the force of gravity pulling it downward, and the force of the rope pulling it upward. The net force on the block is the difference between these two forces, and it determines the acceleration of the block. If the acceleration is 3.6 m/s², then we know that the net force on the block is:

Fnet = m * a

[tex]Fnet = 3.4 kg * 3.6 m/s²[/tex]

Fnet = 12.24 N

Now, we can find the force exerted by the rope on the block by adding the force of **gravity **to the net force and taking the absolute value (since the force of gravity and the force of the rope are in opposite directions):Frope = |Fnet + Fg|where:Frope = force exerted by the rope on the blockFnet = net force on the blockFg = force of gravity on the blockFg = m * gwhere:g = acceleration due to

[tex]gravity = 9.81 m/s²F[/tex]

[tex]g = 3.4 kg * 9.81 m/s²F[/tex]

g = 33.354 N

Now, we can substitute the values we have calculated to get:

Frope = |12.24 N + 33.354 N|Frope = 45.594 NThe **force **that the rope exerts on the block of mass 3.4 kg when it is pulled up off the ground with an acceleration of [tex]3.6 m/s²[/tex] is 45.594 N. Rounded to two decimal places, this is equal to 37.24 N.

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