My midterm is tomorrow and I have a couple questions that I need to study. Ive tried and tried and cannot figure these out.
1) A ball is dropped from a roof and takes 3.0s to reach the ground. Calculate the elevation of the roof above the ground in meters. (neglect air resistance)
2) A car with a mass of 1500-kg changes its speed from 10.0 m/s to 30.0 m/s during a 10.0s- interval. Calculate the net average force acting on the car during the 10.0s-interval.
1) d= 1/2 a t^2 solve for t.
2) a= deltaV/time
but force= m*a = m * above.
1) To calculate the elevation of the roof above the ground, you can use the equations of motion. In this case, the ball is dropped from rest, so its initial velocity is 0 m/s. The acceleration due to gravity is approximately 9.8 m/s^2.
The equation you can use is:
s = ut + (1/2)at^2
Where:
s = displacement (elevation of the roof above the ground)
u = initial velocity (0 m/s)
a = acceleration due to gravity (-9.8 m/s^2, considering downward as negative)
t = time taken to reach the ground (3.0 s)
Plugging in the values, the equation becomes:
s = (0) * 3.0 + (1/2) * (-9.8) * (3.0)^2
Simplifying this further, you get:
s = - (4.9) * (3.0)^2
s = - (4.9) * 9.0
s = - 44.1 meters
The elevation of the roof above the ground is approximately -44.1 meters. The negative sign indicates that the displacement is in the downward direction.
2) To calculate the net average force acting on the car during the 10.0s interval, you can use Newton's second law of motion. The equation for force is:
F = m * a
Where:
F = force (net average force acting on the car)
m = mass of the car (1500 kg)
a = acceleration (change in velocity divided by time interval)
The change in velocity is given as 30.0 m/s - 10.0 m/s = 20.0 m/s. The time interval is given as 10.0 s. Plugging these values into the equation:
F = 1500 kg * (20.0 m/s / 10.0 s)
Simplifying further, you get:
F = 1500 kg * 2.0 m/s^2
F = 3000 N
The net average force acting on the car during the 10.0s interval is 3000 Newtons.