The scheme whereby occupants in a pair of shuttles is as follows
use a strong cable with large weight on the end
Then use the orbital naneuvering system(OMS) to set the whole work as spinning about their common center of gravity.
At a given moment in time, the instantaneous speed can be thought of as the magnitude of instantaneous velocity.
Instantaneous speed is the magnitude of the instantaneous velocity, the instantaneous velocity has direction but the instantaneous speed does not have any direction. Hence, the instantaneous speed has the same value as that of the magnitude of the instantaneous velocity. It doesn't have any direction.
The work done by friction to move the sled is - 1,323 J.
<h3>
What is Coefficient of friction?</h3>
- The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them.
- Typically, it is represented by the Greek letter µ. In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
- The coefficient of friction has no dimensions because both F and N are measured in units of force (such as newtons or pounds). For both static and kinetic friction, the coefficient of friction has a range of values.
- When an object experiences static friction, the frictional force resists any applied force, causing the object to stay at rest until the static frictional force is removed. The frictional force opposes an object's motion in kinetic friction.
Solution:
Given that
Coefficient of friction (µ) = 0.10
Mass (m) = 90kg
distance covered (d) = 30m
We use the formula:
friction work = -µmgdcos∅
friction work = -0.100 × 90 kg × 9.8 m/s² × 30 m × cos 60°
friction work = - 1,323 J
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Answer:
47.8rad/s
Explanation:
For energy to be conserved.
The potential energy sustain by the object would be equal to K.E
P.E = m× g× h = 2 × 9.81× 3.5= 68.67J
Now K.E = 1/2 × I × (w1^2 - w0^2)
I = 2/3 × M × R2
= 2/3 × 2 × (0.23)^2= 0.0705
Hence
W1 = final angular velocity
Wo = initial angular velocity
From P.E = K.E we have;
68.67J = 1/2 × 0.0705 × (w1^2 - w0^2)
(w1^2 - w0^2) = 1948.09
W1^2 = 1948.09 + (18.3^2)
W1^2=2282.98
W1 = √2282.98
=47.78rad/s
= 47.8rad/s to 1 decimal place.
