Answer: the average speed of the rat from the information given above is 0.7m/s
Explanation:
position is given as
x(t) = pt² + qt
finding the diffencial of x(t) with respect to t, we have
d(x(t))/dt = 2pt + q
we substitute the p = 0.36m/s² and q= -1.10 m/s
d(x(t))/dt = 2(0.36)t + (-1.10)
so, at t= 1s
d(x(t))/dt = 2*(0.36)*1 - 1.1 = 0.72 - 1.1 = -0.38m/s
at t= 4s
d(x(t))/dt = 2*(0.36)*4 - 1.10 = 2.88 - 1.10 = 1.78 m/s
To find the average speed,
average speed = (V1 + V2)/ 2
average speed = (1.78 + (-0.38))/2 = 0.7m/s
Answer:
v2 = 27.3m/s
Explanation:
Assuming forward as positive.
Mass = m1 = 64kg
Let v be the common velocity of the student and the skateboard.
mass of skateboard = m2 = 5.94kg
v = 1.4m/s
Since the skateboard and the student are initially moving together at the same velocity their momentum together is
(m1 + m2)v
Let the final velocity of the student be v1 and the final velocity of the skateboard be v2
v1 = – 1.0m/s (falls backwards that's why the velocity is negative since we are assuming forward as positive)
Then from conservation of momentum, momentum before is equal to momentum after.
(m1 + m2)v = m1v1 + m2v2
m2v2= (m1 + m2)v – m1v1
v2 = ( (m1 + m2)v – m1v1)/m2
v2 = ( (64 + 5.94)×1.4 – 64×(-1.0))/5.94
v2 = ( (64 + 5.94)×1.4 + 64×1.0)/5.94
v2 = 27.3m/s
Answer:
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Answer:
The prediction for its maximum potential energy is 109,375 J
Explanation:
Given;
mass of the coaster car, m = 350 kg
speed of the coaster car at the lowest point, v = 25 m/s
The coaster car will have maximum kinetic energy at the lowest point and based on law of conservation of mechanical energy, the maximum kinetic energy of the coaster car at the lowest point will be equal to maximum potential energy at the highest point.
Therefore, the prediction for its maximum potential energy is 109,375 J
At rest because if the distance is not changing, then it is not moving any further, so it must not be moving! The time keeps going no matter what, so the distance, whether it is 0 m or 10,000 km, if the y is horizontal the distance does not change.
