3. A 40-gram ball of clay is dropped from a height, h, above a cup which is attached to a spring of spring force constant, k, of
1 answer:
Answer:
the maximum speed of the ball is 12.65 m/s
Explanation:
Given;
mass of the ball, m = 40 g = 0.04 kg
spring constant, k = 25 N/m
Apply the principle of conservation of energy;
The Elastic potential energy of the spring will be converted into Kinetic of the ball;
Therefore, the maximum speed of the ball is 12.65 m/s
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Answer:
6.5 m/s
Explanation:
We are given that
Distance, s=100 m
Initial speed, u=1.4 m/s
Acceleration,
We have to find the final velocity at the end of the 100.0 m.
We know that
Using the formula
Hence, her final velocity at the end of the 100.0 m=6.5 m/s
Answer:
The minimum coefficient of friction is 0.27.
Explanation:
To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.
Centripetal force is written as
with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as
with ω denoting the angular velocity, which we are given. With that, the above becomes:
Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?
Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:
and so 45 rev/min = 4.71 rad/s.
A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.