A 20 cm long spring is attached to a wall. The spring stretches to a length of 22 cm when you pull on it with a force of 100 n.
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<u>If the disk turns with constant angular velocity, the following statements about it are true </u>
- The linear acceleration of Q is twice as great as the linear acceleration of P
- is moving twice Q as fast as P.
Answer: Options D and E
<u>Explanation: </u>
Let us consider that R is the radius of the circular disc. So as Q is on the rim, so the distance of Q from the centre of the disc is R and as P is the midpoint between centre and rim of the disk, so the distance of P from the centre is R/2.
As we know that the angular velocity of the circular disk will be equal to the ratio of distance covered by that point to the time taken. So the angular velocity at point Q will be
As R is the distance of point Q from the centre of the disc.
Similarly ,
So if we equate v with v’ we obtain that
Therefore, the point Q will be moving twice as fast as P. As the velocity of Q is more than O, the linear acceleration of point Q will also be twice as great as the linear acceleration of P.
This is because acceleration is directly proportional to the rate of change in velocity. So if velocity increases in the factor of 2, the acceleration of point Q will also increase twice with respect to point P.
Answer:
Explanation:
Given data
Mass m=1.2 g=0.0012 kg
diameter d=0.76 m
Friction Force F=3.6 N
To find
Velocity v
Solution
From the Centripetal force we know that
Where m is mass
v is velocity
r is radius
Substitute the given values to find velocity v
So