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s344n2d4d5 [400]
3 years ago
14

The ball has 7.35 joules of potential energy at position B. At position A, all of the energy changes to kinetic energy. The velo

city of the ball at position A is meters/second. Assume there’s no air resistance. Use g = 9.8 m/s2 , PE = m × g × h, and . mass is 1.5 kilograms and g= 9.8 and the hight of position b is 0.5 meters.
Physics
1 answer:
Lina20 [59]3 years ago
3 0
I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J. 
At point A, all this energy has converted into kinetic energy, which is:
K= \frac{1}{2}mv^2
And since K=7.35 J, we can find the velocity, v:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 7.35 J}{1.5 kg} }=3.1 m/s
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C?

Explanation:

My best guess would be C as it's the only answer that gives a reason behind the statement.

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A pulley with a mechanical advantage of 5 will require you to pull _____ times the amount of rope.
Marat540 [252]

Answer:

It will require a force of 1/5, answer A.

Explanation:

In the attached image we can see an example of an array of pulleys that will lift a 100 kg-f load.

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In the pulley D we have three forces of 20 kg-f each and those forces plus the forces in the pulley B, sum a total of 100 kg-f (60+40). This matches the mechanical advantage (100/5) = 20 kg-f

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2 years ago
An 805-kg race car can drive around an unbanked turn at a maximum speed of 54 m/s without slipping. the turn has a radius of cur
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(a) Equating centripetal force to friction force, one finds the relation
  v² = kar
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There is already downward acceleration due to gravity. The additional accceleration due to the wing is
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8 0
3 years ago
A 0.140 kg baseball is thrown horizontally with a velocity of 28.9 m/s. It is struck with a constant horizontal force that lasts
Viefleur [7K]

Answer:

4987N

Explanation:

Step 1:

Data obtained from the question include:

Mass (m) = 0.140 kg

Initial velocity (U) = 28.9 m/s

Time (t) = 1.85 ms = 1.85x10^-3s

Final velocity (V) = 37.0 m/s

Force (F) =?

Step 2:

Determination of the magnitude of the horizontal force applied. This can be obtained by applying the formula:

F = m(V + U) /t

F = 0.140(37+ 28.9) /1.85x10^-3

F = 9.226/1.85x10^-3

F = 4987N

Therefore, the magnitude of the horizontal force applied is 4987N

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