A boy throws a steel ball straight up. consider the motion of the ball only after it has left the boy's hand but before it touch
2 answers:
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In this problem we have the electric field intensity E:
E = 6.5 × newtons/coulomb
We have the magnitude of the load:
q = 6.4 × coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 × meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 × )(6.5 ×
)(1.2 ×
)
PE = 5.0 x joules
None of the options shown is correct.
Answer:
Speed at which the ball passes the window’s top = 10.89 m/s
Explanation:
Height of window = 3.3 m
Time took to cover window = 0.27 s
Initial velocity, u = 0m/s
We have equation of motion s = ut + 0.5at²
For the top of window (position A)
For the bottom of window (position B)
We also have
Solving
So after 1.11 seconds ball reaches at top of window,
We have equation of motion v = u + at
Speed at which the ball passes the window’s top = 10.89 m/s
Answer:
the velocity of the water flow is 7.92 m/s
Explanation:
The computation of the velocity of the water flow is as follows
Here we use the Bernouli equation
As we know that
= 7.92 m/s
Hence, the velocity of the water flow is 7.92 m/s
We simply applied the above formula so that the correct value could come
And, the same is to be considered