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grigory [225]
3 years ago
11

The component of the ball's velocity whose magnitude is most affected by the collisions is

Physics
1 answer:
bazaltina [42]3 years ago
8 0
<span>The component most affected by the collisions is vertical. The ball's vertical will either decrease or increase due to the collision. If the velocity is high during the collsion the ball's vertical will likely be higher and if the ball's velocity is low the vertical will be as well.</span>
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A motorcycle accelerates uniformly from rest at 7.9\,\dfrac{\text{m}}{\text{s}^2}7.9 s 2 m ​ 7, point, 9, space, start fraction,
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Answer:

t = 3.516 s

Explanation:

The most useful kinematic formula would be the velocity of the motorcylce as a function of time, which is:

v(t) = v_0 +at

Where v_0 is the initial velocity and a is the acceleration. However the problem states that the motorcyle start at rest therefore v_0 = 0

If we want to know the time it takes to achieve that speed, we first need to convert units from km/h to m/s.

This can be done knowing that

1 km = 1000 m

1 h = 3600 s

Therefore

1 km/h = (1000/3600) m/s = 0.2777... m/s

100 km/h = 27.777... m/s

Now we are looking for the time t, for which v(t) = 27.77 m/s. That is:

27.777 m/s = 7.9 m/s^2 t

Solving for t

t = (27.7777 / 7.9) s = 3.516 s

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Papessa [141]
<span>3.78 m Ignoring resistance, the ball will travel upwards until it's velocity is 0 m/s. So we'll first calculate how many seconds that takes. 7.2 m/s / 9.81 m/s^2 = 0.77945 s The distance traveled is given by the formula d = 1/2 AT^2, so substitute the known value for A and T, giving d = 1/2 A T^2 d = 1/2 9.81 m/s^2 (0.77945 s)^2 d = 4.905 m/s^2 0.607542 s^2 d = 2.979995 m So the volleyball will travel 2.979995 meters straight up from the point upon which it was launched. So we need to add the 0.80 meters initial height. d = 2.979995 m + 0.8 m = 3.779995 m Rounding to 2 decimal places gives us 3.78 m</span>
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