Answer:
≈ 6.68 m/s
Explanation:
A suitable formula is ...
vf^2 -vi^2 = 2ad
where vi and vf are the initial and final velocities, a is the acceleration, and d is the distance covered.
We note that if the initial launch direction is upward, the velocity of the ball when it comes back to its initial position is the same speed, but in the downward direction. Hence the problem is no different than if the ball were initially launched downward.
Then ...
vf = √(2ad +vi^2) = √(2·9.8 m/s^2·1.0 m+(5 m/s)^2) = √44.6 m/s
vf ≈ 6.68 m/s
The ball hits the ground with a speed of about 6.68 meters per second.
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We assume the launch direction is either up or down.
Answer:
Speed is the time rate at which an object is moving along a path while velocity is the rate and direction of an object's movement
Answer:
12294.31 m/s
Explanation:
Momentum = (mass)(velocity)
Momentum before = Momentum after
(momentum of bullet)+(momentum of block)=(momentum of bullet and block)
0.035v+50(0)=(0.035+50)(8.6)
0.035v=430.301
v=12294.3142857m/s
Answer:
It takes 11.20 ms for the sphere to increase in potential by 1260 V
Explanation:
Using the formula q = It where I is current in Ampere, and t is in seconds
q = ( 1.000002 - 1.0000000) t
q = 0.000002t
Voltage on the surface of the sphere V = Kq / r
where K is 9.0 × 10⁹ N².m² / c² and
R = 16 cm = 16 /100 = 0.16m
V = Kq /r
substitute the value into q
V = K(0.000002t) / r
cross multiply
rV = K × 0.000002t
make t subject of the formula
t = 0.16 × 1260 / ( 9×10⁹ × 0.000002)
t = 201.6 / ( 18 × 10³)
t = 0.0112 s
= 11.20 ms
It takes 11.20 ms for the sphere to increase in potential by 1260 V