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yuradex [85]
2 years ago
7

Car B is being pushed by a force of 22000 N. If it has a mass of 1375 kg.,

Physics
1 answer:
Burka [1]2 years ago
5 0

Answer:

a = 16 m/s²

General Formulas and Concepts:

<u>Dynamics</u>

Newton's Law of Motions

  • Newton's 1st Law of Motion: An object at rest remains at rest and an object in motion stays in motion
  • Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration)
  • Newton's 3rd Law of Motion: For every action, there is an equal and opposite reaction<u> </u>

Explanation:

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] F = 22000 N

[Given] m = 1375 kg

[Solve] a

<u>Step 2: Find Acceleration</u>

  1. Substitute in variables [Newton's 2nd Law of Motion]:                                 22000 N = (1375 kg)a
  2. Isolate <em>a</em>:                                                                                                            16 m/s² = a
  3. Rewrite:                                                                                                             a = 16 m/s²
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Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

V_{Ax}=V_{A}cos\theta

V_{Ax}=(21m/s)cos(-14^{o})

V_{Ax}=20.38 m/s

V_{Ay}=V_{A}sin\theta

V_{Ay}=(21m/s)sin(-14^{o})

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we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

x_{A}=V_{Ax}t

x_{A}=(20.38m/s)(0.317s)

x_{A}=6.46m

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

V_{Bx}=20.88 m/s

V_{By}=-2.195 m/s

y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}

0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}

-4.9t^{2}-2.195t+2.1=0

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}

t= -0.9159s    or   t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

x_{B}=V_{Bx}t

x_{B}=(20.88m/s)(0.468s)

x_{B}=9.77m

So once we got the two distances we can now find the difference between them:

x_{B}-x_{A}=9.77m-6.46m=3.31m

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

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