The diagram below shows the velocity vectors for two cars that are moving
relative to each other.
45 m/s west
25 m/s east
Car 1
Car 2
From the frame of reference of car 1, what is the velocity of car 2?
OA. 20 m/s west
OB. 70 m/s east
C. 70 m/s west
D. 20 m/s east

a. To find the acceleration, we can use the equation:

acceleration = change in velocity / time taken

Here, the initial velocity is 0 m/s (since she starts from rest), the final velocity is 5 m/s, and the time taken is 10 s. Therefore:

acceleration = (5 m/s - 0 m/s) / 10 s = 0.5 m/s^2

So the acceleration of the girl on her bicycle is 0.5 m/s^2.

b. The average speed during the 10 s can be found by dividing the total distance traveled by the time taken. We don't know the distance traveled yet, so we can use another equation:

distance = (initial velocity x time taken) + (0.5 x acceleration x time taken^2)

Here, the initial velocity is 0 m/s, and the time taken is 10 s. We already calculated the acceleration in part (a) as 0.5 m/s^2. Plugging these values in, we get:

distance = (0 m/s x 10 s) + (0.5 x 0.5 m/s^2 x (10 s)^2) = 25 meters

So the girl traveled 25 meters in 10 seconds.

c. As the girl pedals, she applies force to the pedals, which in turn transfers the force to the rear wheel. This force drives the bicycle forward, causing it to gain speed. However, when the bicycle picks up speed, air resistance (also called drag) comes into play, which increases. Eventually, the force of air resistance becomes equal and opposite to the force that propels the bicycle forward. This means that the net force on the bicycle becomes zero, stopping its acceleration and reaching its maximum speed, known as the terminal velocity. Even if the girl pedals faster, she won't be able to increase her speed because the drag force counteracts the propelling force.

During Simple Harmonic Motion (SHM), an object at the mean position has potential energy (P.E)=0 and kinetic energy (K.E)=MAX. So, the correct option is (d).

In Simple Harmonic Motion (SHM), an object oscillates about a mean position, with the motion characterized by a restoring force proportional to its displacement from the mean position.

When the object is at the mean position, it has maximum kinetic energy (K.E) because it is at its maximum velocity, and it has zero potential energy (P.E) since it is not displaced from the equilibrium position.

As the object moves further from the mean position, its P.E increases, and K.E decreases. The correct answer to the question is option (d), where P.E=0 and K.E=MAX at the mean position.

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The magnitude of the resultant vector is obtained by finding the sum of two horizontal vectors and the sum of two vertical vectors. Thus, option C is correct.

The resultant vector is the single vector that has the same effect in the number of vectors collectively produced. The resultant vector in the horizontal and vertical direction is obtained by drawing a diagonal and hence by using the Pythagoras theorem.

To find resultant vector is obtained by finding the sum of two horizontal vectors and vertical vectors and then using the Pythagoras theorem.

Thus, the ideal solution is option C.

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Explanation: ΔL = τ(average) * Δt
Change in angular momentum = average torque * change in time

solve for average torque for each objects
τ(average) = ΔL / Δt

Object y average torque
τy = ΔLy / Δt = 20 / 5 = 4
τy = 4

Object x average torque
τx = ΔLx / Δt = 10 / 5 = 2
τx = 2

Relates τy and τx
2τx = τy

Two objects, X and Y, experience external net torques that vary over a period of 5 seconds. Object X has a moment of inertia I0, and Object Y has a moment of inertia 2I0. The average value of the magnitude of the external net torque exerted on Object X from time t=0 to t=5s is torquex. Similarly, the average value for ObjectY is torquey.

The magnitudes of the angular momenta L of Objects X and Y versus t are shown in the graph. The precise relation between torquey and torquex is torquey = 2 * torquex.

To relate torquey to torquex, we are able to use the concept of angular momentum and torque. Angular momentum is described because the manufactured from the moment of inertia and angular velocity:

L = I * ω

Differentiating this equation with an appreciation of time, we get:

dL/dt = d(I * ω)/dt

Using the product rule of differentiation, we've got:

dL/dt = I * dω/dt + ω * dI/dt

Now, we realize that torque (τ) is described because of the charge of the exchange of angular momentum:

τ = dL/dt

Substituting the expression for dL/dt in terms of angular velocity and second of inertia:

τ = I * dω/dt + ω * dI/dt

Let's denote the common price of torque for item X as torquex. Since object X has a moment of inertia I0, we can write:

torquex = I0 * dω/dt + ω * dI0/dt

Now, let's consider item Y. It has a moment of inertia 2I0. Using the identical expression, we will write:

torquey = (2I0) * dω/dt + ω * d(2I0)/dt

torquey = 2I0 * dω/dt + ω * (2 * dI0/dt)

torquey = 2I0 * dω/dt + 2ω * dI0/dt

Comparing the expressions for torquex and torquey, we will see that:

torquey = 2 * torquex

Therefore, the precise relation between torquey and torquex is;

torquey = 2 * torquex.

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The correct question is;

"Two objects, X and Y, experience external net torques that vary over a period of 5 seconds. Object X has a moment of inertia I0, and Object Y has a moment of inertia 2I0. The average value of the magnitude of the external net torque exerted on Object X from time t=0 to t=5s is torquex. Similarly, the average value for ObjectY is torquey. The magnitudes of the angular momenta L of Objects X and Y versus t are shown in the graph. Which of the following expressions correctly relates torquey to torquex?"

Move numbers to the boxes to show the factor pairs of 14:

Blank space 1: 1

Blank space 2: 1

Blank space 3: 2

Blank space 4: 7

A factor pair of a number is a pair of whole numbers that can be multiplied together to give the original number. For the number 14, the factor pairs are (1,14) and (2,7). So, we can put 1 in the first blank, 14 in the second blank, 2 in the third blank, and 7 in the fourth blank to show the factor pairs of 14.

To show the factor pairs of 14 in the given response area with 4 blank spaces, we need to find the two numbers that can be multiplied together to give 14. These two numbers are called factor pairs of 14.

To begin, we can start listing the factors of 14. The factors of 14 are 1, 2, 7, and 14. We can then use these factors to form factor pairs by multiplying them together. The factor pairs of 14 are (1, 14) and (2, 7).

To show these factor pairs in the given response area, we can put the first factor of each pair in the first and third blank spaces, and the second factor of each pair in the second and fourth blank spaces. Therefore, we can put 1 in the first blank, 14 in the second blank, 2 in the third blank, and 7 in the fourth blank to show the factor pairs of 14.

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The amount of acetone that was spilt is around 0.36 liters, or 360 milliliters.

To solve this problem, use the formula for volumetric thermal expansion:

ΔV = V₀βΔT

Where:

ΔV = change in volume

V₀ = initial volume

β = coefficient of volumetric expansion

ΔT = change in temperature

Also use the formula for linear thermal expansion to find the change in length of the container:

ΔL = L₀αΔT

Where:

ΔL = change in length

L₀ = initial length

α = coefficient of linear expansion

ΔT = change in temperature

Initial length can be calculated as follows:

V = L³ ⇒ L = ∛V = ∛6 L ≈ 1.82 meters

Now calculate the change in length of the container and the change in volume of acetone:

ΔL = L₀αΔT = (1.82 m)(10⁻⁵ °C⁻¹)(40 °C) ≈ 0.00073 meters

ΔV = V₀βΔT = (6 liters)(1.5 x 10⁻³ °C⁻¹)(40 °C) ≈ 0.36 liters

Since the acetone spills out of the container, its final volume is equal to the initial volume minus the change in volume:

Vf = Vi - ΔV = 6 L - 0.36 L = 5.64 L

Therefore, the volume of spilled acetone is approximately 0.36 liters or 360 milliliters.

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The best example of Newton's third law of motion is, When a passenger stepped from a boat to the shore, the boat moved away from the shore. Thus, option C is correct.

Sir Issac Newton gives three laws of motion. The first law states that an object remains at rest or in continuous motion unless an external force acted on it. The second law stated that the force is directly proportional to the acceleration of the object. Newton's third law states that, for every action, there is an equal and opposite reaction.

From the given, Newton's third law is applicable, When a passenger stepped from a boat to the shore, the boat moved away from the shore. This shows the action and reaction of the boat and shore.

Thus, the ideal solution is option C.

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The reflection in a clear window of a store is a(n) image.

Images are seen as reflections because of the behavior of light. When light strikes a smooth, reflective surface such as a mirror or still water, it bounces off the surface at the same angle at which it hit it. This process is called reflection. The reflected light rays then travel to our eyes, creating an image.

The angle of incidence (the angle at which the light strikes the surface) is equal to the angle of reflection (the angle at which the light bounces off the surface). This causes the reflected image to be a mirror image of the original object.

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The skier was initially going at 8.046 m/s velocity.

Assume that the skier's initial velocity is v.

The skier travelled 22 metres before coming to a stop.

The kinetic friction coefficient between the skier and the snow is 0.15.

The force of friction acting on the skier is given by:

friction = coefficient of friction multiplied by normal force

The normal force is equal to the weight of the skier, which can be calculated as:

weight = mass x gravity

Assuming the mass of the skier to be m and the acceleration due to gravity to be g, we get:

weight = m x g

The force of friction can then be calculated as:

friction = 0.15 x m x g

The work done by the force of friction is equal to the initial kinetic energy of the skier, which can be expressed as:

work done by friction = 0.5 x m x [tex]v^2[/tex]

Equating the work done by friction to the initial kinetic energy, we get:

0.5 x m x [tex]v^2[/tex] = 0.15 x m x g x 22

Simplifying the equation, we get:

[tex]v^2[/tex] = 2 x 0.15 x 9.8 x 22

When we take the square root of both sides, we get:

v = 8.046 m/s

As a consequence, the skier began moving at 8.046 m/s.

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The work done is equal to the force applied parallel to the floor multiplied by the distance moves. Hence, option B is correct.

When a wooden crate is pushed across a floor, the work done on it is given by the product of the force applied parallel to the floor and the distance moved by the crate.

This is because work is defined as the product of force and displacement in the direction of the force, which in this case is the distance moved by the crate. The force of friction and the mass of the crate are not relevant to the calculation of work done in this scenario

It is important to understand the concept of work and the factors that influence it, as it is a fundamental concept in physics and is used in many real-world applications.

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Answer:

The correct answer is B. Energy cannot be created or destroyed.

The law of conservation of energy states that energy cannot be created or destroyed, but it can be transformed from one form to another. This means that in any physical process, the total amount of energy in a system remains constant. Energy can be converted from one form to another, such as from kinetic energy to potential energy or from electrical energy to light energy, but the total amount of energy in the system remains the same.