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Neporo4naja [7]
3 years ago
11

A 1500 kg car

Physics
1 answer:
Lera25 [3.4K]3 years ago
5 0
<h2>We determine the final velocity, </h2><h2>v</h2><h2>f</h2><h2>, of the car. We do this by applying the kinematic equation,</h2><h2 /><h2>v</h2><h2>2</h2><h2>f</h2><h2>=</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>a</h2><h2>d</h2><h2 /><h2>where </h2><h2>v</h2><h2>i</h2><h2> is the initial velocity, a is the acceleration, and d is the distance covered across the movement. We can express the acceleration using the equation,</h2><h2 /><h2>a</h2><h2>=</h2><h2>F</h2><h2>m</h2><h2 /><h2>where F is the applied force and m is the mass of the car. We use the following values for the variables:</h2><h2 /><h2>v</h2><h2>i</h2><h2>=</h2><h2>5.25</h2><h2> </h2><h2>m</h2><h2>s</h2><h2>−</h2><h2>1</h2><h2>F</h2><h2>=</h2><h2>1250</h2><h2> </h2><h2>N</h2><h2>m</h2><h2>=</h2><h2>1500</h2><h2> </h2><h2>k</h2><h2>g</h2><h2>d</h2><h2>=</h2><h2>42.8</h2><h2> </h2><h2>m</h2><h2>We proceed with the solution.</h2><h2 /><h2>v</h2><h2>2</h2><h2>f</h2><h2>=</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>a</h2><h2>d</h2><h2>v</h2><h2>2</h2><h2>f</h2><h2>=</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>F</h2><h2>d</h2><h2>m</h2><h2>Take the positive root.</h2><h2>v</h2><h2>f</h2><h2>=</h2><h2>√</h2><h2>v</h2><h2>2</h2><h2>i</h2><h2>+</h2><h2>2</h2><h2>F</h2><h2>d</h2><h2>m</h2><h2>=</h2><h2>√</h2><h2>(</h2><h2>5.25</h2><h2> </h2><h2>m</h2><h2>s</h2><h2>−</h2><h2>1</h2><h2>)</h2><h2>2</h2><h2>+</h2><h2>2</h2><h2>(</h2><h2>1250</h2><h2> </h2><h2>N</h2><h2>)</h2><h2>(</h2><h2>42.8</h2><h2> </h2><h2>m</h2><h2>)</h2><h2>1500</h2><h2> </h2><h2>k</h2><h2>g</h2><h2>≈</h2><h2>9.9</h2><h2> </h2><h2>m</h2><h2>s</h2><h2>−</h2><h2>1</h2><h2>Help improve Study.com. Report an Error</h2>

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A force of 10 N acts on an object with a mass of 10 kg. What is its acceleration? A. 1 m/s2 B. 10 m/s2 C. 2 m/s2 D. 5 m/s2
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The acceleration exerted by the object of mass 10 kg is \mathbf{1} m / \boldsymbol{s}^{2}

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<u>Explanation:</u>

According to Newton’s second law of motion, any external force acting on a body will be directly proportional to the mass of the body as well as acceleration exerted by the body. So, the net external force acting on any object will be equal to the product of mass of the object with acceleration exerted by the object. Thus,

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So,

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As the force acting on the object is stated as 10 N and the mass of the object is given as 10 kg, then the acceleration will be

                  Acceleration =\frac{10 \mathrm{N}}{10 \mathrm{kg}}=1 \mathrm{m} / \mathrm{s}^{2}

So, the acceleration exerted by the object of mass 10 kg is \mathbf{1} m / \boldsymbol{s}^{2}

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A stone is thrown vertically upwards with an initial velocity of 20m/sec. Find the maximum height ot reaches and the time taken
MAXImum [283]

Answer:

The height reached is 20m, The time taken to reach 20m is 2 seconds

Explanation:

Observing the equations of motion we can see that the following equation will be most helpful for this question.

v^{2} = u^{2} + 2as

We are given initial velocity, u

We know that the stone will stop at its maximum height, so final velocity, v

Acceleration, a

And we are looking for the displacement (height reached), s

Substitute the values we are given into the equation

0^{2} = 20^{2} + 2(10)s

Rearrange for s

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s = -20 (The negative is just showing direction, it can be ignored for now)

The height reached is 20m

Use a different equation to find the time taken

s = vt - \frac{1}{2} at^{2}

Substitute in the values we have

-20=(0)t - \frac{1}{2} (10)t^{2}

Rearrange for t

-20 =0 -5 t^{2}

\frac{-20}{-5} =t^{2}

4 = t^{2}

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The time taken to reach 20m is 2 seconds

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