The displacement and velocity of a rock that is dropped from rest after 4s, is 6 km/h. This can vary depending on the speed of the rock, and the surroundings.
After you dropped a rock in a cup of water you noticed some displacement of the water on the counter.
Any object which is at rest has zero velocity, for example a rock on the road. car parked at lane, a motor which is not moving or stop. a man sleeping. home office statue of liberty. anything that is at rest has zero velocity.
Since the truck is moving at constant velocity, the rock will also fall straight down due to gravity from the perspective of an observer inside the truck. The horizontal motion of the truck does not affect the vertical motion of the falling rock. Therefore, the rock will follow a vertical straight-line path and hit the floor directly below where it was dropped.
To find the velocity of the rock when it hits the ground, we can use the equation of motion: ( v = u + at ), where ( v ) is the final velocity, ( u ) is the initial velocity (which is 0 m/s since it falls from rest), ( a ) is the acceleration due to gravity (approximately 9.81 m/s(^2)), and ( t ) is the time taken (1.5 s). Plugging in the values, we get ( v = 0 + 9.81 \times 1.5 = 14.715 , \text{m/s} ). Therefore, the velocity of the rock when it hits the ground is 14.715 m/s.
The acceleration of the rock is calculated by dividing the change in velocity by the time taken. In this case, the change in velocity is 4.9 m/s and the time taken is 3 seconds. Thus, the acceleration of the rock is 4.9 m/s^2.
Yes
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After you dropped a rock in a cup of water you noticed some displacement of the water on the counter.
A truck moving at constant velocity inside the storage compartment a rock is dropped from the midpoint of the ceiling and strikes the floor below the rock hits the floor Read more:
S=ut+0.5*a*t2 s=displacement u=intial velocity=0m/s t=time=7seconds a=9.81m/s2 s=0*7+0.5*9.81*72 s=240.345m
It will have both horizontal and vertical velocity...think about it, if you were said bird flying through the sky at say 35 mph, and you dropped a rock then the rock would fall, but it would still be moving forward and it would fall the same way a baseball falls after it reaches the top of the throw.
The process in which sediment moved y erosin is dropped and comes to rest
Any object which is at rest has zero velocity, for example a rock on the road. car parked at lane, a motor which is not moving or stop. a man sleeping. home office statue of liberty. anything that is at rest has zero velocity.
Since the truck is moving at constant velocity, the rock will also fall straight down due to gravity from the perspective of an observer inside the truck. The horizontal motion of the truck does not affect the vertical motion of the falling rock. Therefore, the rock will follow a vertical straight-line path and hit the floor directly below where it was dropped.
To find the velocity of the rock when it hits the ground, we can use the equation of motion: ( v = u + at ), where ( v ) is the final velocity, ( u ) is the initial velocity (which is 0 m/s since it falls from rest), ( a ) is the acceleration due to gravity (approximately 9.81 m/s(^2)), and ( t ) is the time taken (1.5 s). Plugging in the values, we get ( v = 0 + 9.81 \times 1.5 = 14.715 , \text{m/s} ). Therefore, the velocity of the rock when it hits the ground is 14.715 m/s.
The acceleration of the rock is calculated by dividing the change in velocity by the time taken. In this case, the change in velocity is 4.9 m/s and the time taken is 3 seconds. Thus, the acceleration of the rock is 4.9 m/s^2.
The acceleration of the rock would be (1.63 , \text{m/s}^2) (calculated by dividing the change in velocity by the time taken).