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.
Answer:
The measured redshift is z =2
Explanation:
Since the object is traveling near light speed, since v/c = 0.8, then we have to use a redshift formula for relativistic speeds.
Finding the redshift.
We can prepare the formula by dividing by lightspeed inside the square root to both numerator and denominator to get
Replacing the given information
Thus the measured redshift is z = 2.
Answer:
a) KE = 888.26J
b) N = 294.5 turns
Explanation:
For the kinetic energy:
The inertia is:
So, the kinetic energy will be:
Now, friction force is:
Ff = μ*N = 0.80*5N = 4N
The energy balance would be:
Kf - Ko = Wf where Kf=0; Ko = 888.26J; and Wf is the work done by friction force.
Wf = -Ff*d = -Ff*N*2*π*R where N is the amount of turns it gives.
Replacing these values into the energy balance:
0-888.26=-4*N*2*π*0.12
-888.26=-0.96*π*N
N=294.5 turns