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Juliette [100K]
3 years ago
5

A ball is attached to a vertical spring. The ball is initially supported at a height y so that the spring is neither stretched n

or compressed. The ball is then released from rest and it falls to a height y - h before moving upward. Consider the following quantities: translational kinetic energy, gravitational potential energy, elastic potential energy. When the ball was at a height y - (h/2), which of the listed quantities has (have) values other than zero joules
Physics
1 answer:
scoundrel [369]3 years ago
3 0

Answer:

All the three quantities will have non zero joules.

Explanation:

At the initial position of rest the system will have only gravitational potential energy while the other 2 quantities will be zero.

when the system reaches a height (y-h) only kinetic energy will be zero while the other 2 quantities will be non zero

At the position of (y-h/2) all the 3 quantities will be non zero.

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I'm not sure, I think it's option A.

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3 years ago
A cannon is fired straight up into the air. If the cannon ball comes back down to the launch point in 5 seconds, what was the ma
Nana76 [90]

Answer:

30.63 m

Explanation:

From the question given above, the following data were obtained:

Total time (T) spent by the ball in air = 5 s

Maximum height (h) =.?

Next, we shall determine the time taken to reach the maximum height. This can be obtained as follow:

Total time (T) spent by the ball in air = 5 s

Time (t) taken to reach the maximum height =.?

T = 2t

5 = 2t

Divide both side by 2

t = 5/2

t = 2.5 s

Thus, the time (t) taken to reach the maximum height is 2.5 s

Finally, we shall determine the maximum height reached by the ball as follow:

Time (t) taken to reach the maximum height = 2.5 s

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =.?

h = ½gt²

h = ½ × 9.8 × 2.5²

h = 4.9 × 6.25

h = 30.625 ≈ 30.63 m

Therefore, the maximum height reached by the cannon ball is 30.63 m

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3 years ago
Take a look at the weather map. The front seen there causes short periods of storms and heavy rains. What type of front is this?
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I am pretty sure it is a cold front.
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A car traveling at 30 m/s speeds up to 35 m/s over a period of 5 seconds. What is the acceleration of the car?
Delvig [45]

Answer:

u =30 m/s

v = 35 m/s

t = 5 secs

Explanation:

a = (v- u)/t

a = (35-30)/5

a = 5/5

a = 1 m/s^2

PLS MARK BRAINLIEST

8 0
2 years ago
Can you do it for me pls thank and if you do I mark you brainlies
Art [367]

Answer:

800pa

Explanation:

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So there are a wats up group where you can get help step by step and well explained by the trusted experts.

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3 years ago
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