Assuming that it continues to accelerate at the same rate it will take another 10 seconds to reach 40 m/s.
Answer:
Explanation:
Since the first question states that there is a change in the velocity from rest to 20 m/s in 10 seconds time interval. So the acceleration experienced by the car during this 10 seconds should be determined first as follows:
Acceleration = (final velocity-initial velocity)/Time
Acceleration = (20-0)/10 = 2 m/s².
So this means the car is traveling with an acceleration of 2 m/s².
As it is stated that the car continues to move with same acceleration, then in the second case, the acceleration is fixed as 2 m/s², initial velocity as 20 m/s and final velocity as 40 m/s. So the time taken for the car to reach this velocity with the constant acceleration value will be as follows:
Time = Change in velocity/Acceleration
Time = (40-20)/2 = 20/2=10 s
So again in another 10 seconds by the car to reach 40 m/s from 20 m/s. Similarly the car will take a total of 20 seconds to reach from rest to 40 m/s value for velocity.
Decreasing Pressure. The combined gas law states that the pressure of a gas is inversely related to the volume and directly related to the temperature. If temperature is held constant, the equation is reduced to Boyle's law. Therefore, if you decrease the pressure of a fixed amount of gas, its volume will increase.
Explanation:
Given that,
Initial speed of the bag, u = 7.3 m/s
Height above ground, s = 24 m
We need to find the speed of the bag just before it reaches the ground. It can be calculated using third equation of motion as :
v = 22.88 m/s
So, the speed of the bag just before it reaches the ground is 22.38 m/s. Hence, this is the required solution.
Move the objects faster to get more friction.
