4. We have 4 identical strain gauges of the same initial resistance (R) and the same gauge factor (GF). They will be used as R1,
1 answer:
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Answer:
The acceleration is about 9.8 m/s2 (down) when the ball is falling.
Explanation:
The ball at maximum height has velocity zero
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.8 m/s² (positive downward and negative upward)
The accleration 9.8 m/s² will always be acting on the body in opposite direction when the body is going up and in the same direction when the body is going down. The acceleration on the body will never be zero
Answer:
to overcome the out of friction we must increase the angle of the plane
Explanation:
To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.
X axis
fr - Wₓ = m a (1)
Y axis
N- = 0
N = W_{y}
let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
the friction force has the formula
fr = μ N
fr = μ Wy
fr = μ mg cos θ
from equation 1
at the point where the force equals the maximum friction force
in this case the block is still still so a = 0
F = fr
F = (μ mg) cos θ
We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.
This is the force that balances the friction force, any force slightly greater than F initiates the movement.
Consequently, to overcome the out of friction we must increase the angle of the plane
the correct answer is to increase the angle of the plane
Answer:
Explanation:
We can use the conservation of the angular momentum.
Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.
So we will have:
Now, we just need to solve it for ω.
I hope it helps you!