It should be noted that the student should use K = 1/2mv² with the initial speed of the block for one trial.
<h3>
Method of using the data collected.</h3>
From the complete information, the student wants to use the data collected and the known quantities from the experiment to determine the initial total mechanical energy of the block-ramp-Earth system for all trials in the experiment.
In this case, it's important to use K = 1/2mv² with the block's initial speed for one trial due to the fact that the initial speed is the same in all the trials.
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Answer:
The friction force acting on the object is 7.84 N
Explanation:
Given;
mass of object, m = 4 kg
coefficient of kinetic friction, μk = 0.2
The friction force acting on the object is calculated as;
F = μkN
F = μkmg
where;
F is the frictional force
m is the mass of the object
g is the acceleration due to gravity
F = 0.2 x 4 x 9.8
F = 7.84 N
Therefore, the friction force acting on the object is 7.84 N
Answer:
460.52 s
Explanation:
Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that
dV/dt ∝ V
dV/dt = kV
separating the variables, we have
dV/V = kdt
integrating both sides, we have
∫dV/V = ∫kdt
㏑(V/V₀) = kt
V/V₀ = 
Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V
Since dV/dt = kV
-0.01V = kV
k = -0.01
So, V/V₀ = 
V = V₀
Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀
So, V = 0.1V₀
Thus
V = V₀
0.1V₀ = V₀
0.1V₀/V₀ = 
0.1 = 
to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have
㏑(0.01) = -0.01t
So, t = ㏑(0.01)/-0.01
t = -4.6052/-0.01
t = 460.52 s
Answer:
1.3823 rad/s
20.7345 m/s
28.66129935 m/s²

2006.29095 N radially outward
Explanation:
r = Radius = 15 m
m = Mass of person = 70 kg
g = Acceleration due to gravity = 9.81 m/s²
Angular velocity is given by

Angular velocity is 1.3823 rad/s
Linear velocity is given by

The linear velocity is 20.7345 m/s
Centripetal acceleration is given by

The centripetal acceleration is 28.66129935 m/s²
Acceleration in terms of g


Centripetal force is given by

The centripetal force is 2006.29095 N radially outward
The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.