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1 year ago

A small car is pushing a large truck. They are speeding up.

Physics

1 answer:

Gnom [1K]1 year ago

3 0

The force of the truck on the car is equal to the force of the car on the truck.

<h3>What is Newton's third law of motion?</h3>

Newton's third law of motion states that action and reaction are equal and opposite.

Let the mass of the small car = m

Let the mass of the truck = M

The force exerted by the small car is calculated as;

F₁ = ma

The force exerted by the large truck is calculated as;

F₂ = Ma

According to Newton's third law, the magnitude of the two forces are equal but opposite in direction.

|F₁| = |F₂|

Thus, we can conclude that, the force of the truck on the car is equal to the force of the car on the truck.

Learn more about Newton's third law of motion here: brainly.com/question/25998091

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Diamagnetic materialsA) have small negative values of magnetic susceptibility.B) are those in which the magnetic moments of all

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Answer:

C) experience a small induced magnetic moment when placed in an external magnetic field.

Explanation:

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2 years ago

How far does a plane fly in 10.0 s while its velocity is changing from 175 m/s to 105 m/s at a uniform rate of

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Answer:

1400 m

Explanation:

Given:

v₀ = 175 m/s

v = 105 m/s

t = 10.0 s

Find: Δx

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Δx = ½ (105 m/s + 175 m/s) (10.0 s)

Δx = 1400 m

6 0

2 years ago

Read 2 more answers

During a very quick stop, a car decelerates at 7.00 m/s2 . (a) What is the angular acceleration of its 0.280-m-radius tires, ass

love history [14]

Answer:

-25 rad/s²

29 times

3.8 seconds

50.54 m

26.6 m/s

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

a = Acceleration = -7 m/s² (negative because of deceleration)

r = Radius of wheel = 0.28 m

Angular acceleration is given by

$\alpha=\frac{a}{r}\\\Rightarrow \alpha=\frac{-7}{0.28}\\\Rightarrow \alpha=-25\ rad/s^$

Angular acceleration of the wheel is -25 rad/s²

$\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-95}{-25}\\\Rightarrow t=3.8\ s$

It took 3.8 seconds for the car to stop

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=95\times 3.8+\frac{1}{2}\times -25\times 3.8^2\\\Rightarrow \theta=180.5\ rad$

$\frac{180.5}{2\pi}=28.72746\ rev$

The wheel rotated 29 times.

The

Initial velocity

$u=r\omega_i\\\Rightarrow u=0.28\times 95\\\Rightarrow u=26.6\ m/s$

The car’s initial velocity is 26.6 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=26.6\times 3.8+\frac{1}{2}\times -7\times 3.8^2\\\Rightarrow s=50.54\ m$

Distance covered in the time is 50.54 m

If the distance is found in meters we get 50.54 m which is very low. Considering the initial velocity is 95.76 km/h. Coming to a stop in that distance is very low. So, the values do not seem reasonable. At least a 100 m is required to stop from that speed.

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2 years ago

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