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Why does the acceleration of a merry go round change even though the speed stays the same?
The acceleration of a merry-go-round changes because acceleration measures a change in velocity, not just speed. In the case of the merry-go-round, even though the speed may stay constant, the direction of the velocity is constantly changing as the object moves in a circular path, resulting in acceleration due to the change in direction.
Is it true that A person on a merry go round is constantly accelerating away from the center?
true:apex
What is the solution to the merry-go-round physics problem?
The solution to the merry-go-round physics problem involves understanding centripetal force and acceleration. The centripetal force required to keep an object moving in a circular path on a merry-go-round is provided by friction between the object and the surface of the merry-go-round. This force is directed towards the center of the circle and is equal to the mass of the object times its centripetal acceleration. By calculating the centripetal force and acceleration, one can determine the speed at which the object is moving on the merry-go-round.
Is a merry go round accelerating if it has a constant speed?
Yes, a merry-go-round is accelerating even if it has a constant speed because acceleration includes changes in direction as well as changes in speed. In this case, the merry-go-round is constantly changing direction as it spins around the central axis, so it is undergoing acceleration.
How can you accelerate if your speed does not change?
A change in the vector or direction that you are traveling is considered an acceleration; even if you don't change speed. This would be the case for going around in a merry-go-round, e.g. You are accelerating (changing direction), but the speed is constant (velocity is changing).
A person on a merry go round is constantly accelerating toward the center true or false?
False. A person on a merry-go-round is not constantly accelerating towards the center. The person is experiencing centripetal acceleration, which is directed towards the center of the merry-go-round, but it is not a continuous increase in speed or velocity.
A child riding a merry-go-round is accelerating because the child?
is experiencing a centripetal acceleration directed towards the center of the merry-go-round due to the circular motion. This acceleration constantly changes the direction of the child's velocity, even though the speed may remain constant, resulting in acceleration.
What is the relation between speed-acceleration and time?
acceleration = the change in speed over time, assuming a linear motion. For example, the speed is 5 [m/s] at t = 0 and 100 [m/s] at t = 2 [min]. The average acceleration = (100 - 5) [m/s]/(2*60) [s] = 0.792 [m/s2]. When vectors are involved (a roller coaster ride or merry-go-round), the calculations will be a little more complicated, but the expression is very similar: A = (V2-V1)/(t2-t1), where A = the average acceleration vector; Vi = the velocity vector at time i. For the instantaneous acceleration, A = dV/dt. Please see the related link. =============================
A child on a merry-go-round is accelerating because the child?
is changing direction as the merry-go-round spins. Even though the child may maintain a constant speed, the continuous change in direction results in centripetal acceleration. This acceleration is directed towards the center of the circular path, keeping the child moving in a circular motion. Thus, the child experiences acceleration due to the change in velocity direction.
When a person on merry go round is constantly accelerating through the center?
As the person on the merry-go-round accelerates through the center, they will experience a centripetal force directed towards the center of the merry-go-round. This force is responsible for keeping them moving in a circular path. If the acceleration is constant, the person will feel a consistent pull towards the center as they rotate around.
What is the solution to the physics merry go round problem?
The solution to the physics merry-go-round problem involves using the principles of rotational motion and centripetal force to calculate the acceleration and tension in the ropes holding the riders. By applying the equations of circular motion, one can determine the necessary forces and velocities to keep the riders safely on the merry-go-round.
A moving merry-go-round house is not accelerating True or false?
False. Any change in velocity, including a change in direction, is considered acceleration. Since the merry-go-round house is constantly changing its direction as it moves in a circular path, it is indeed accelerating.
