Which of the following has the fewest calence electrons?
1 answer:
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Answer:
<em>155.80rad/s</em>
Explanation:
Using the equation of motion to find the angular acceleration:
is the final angular velocity in rad/s
is the initial angular velocity in rad/s
is the angular acceleration
t is the time taken
Given the following
Time = 4.1secs
Convert the angular velocity to rad/s
1rpm = 0.10472rad/s
6100rpm = x
x = 6100 * 0.10472
x = 638.792rad/s
Get the angular acceleration:
Recall that:
638.792 = 0 + ∝(4.1)
4.1∝ = 638.792
∝ = 638.792/4.1
∝ = 155.80rad/s
<em>Hence the angular acceleration as the blades slow down is 155.80rad/s</em>
Answer:
v = 0
Explanation:
This problem can be solved by taking into account:
- The equation for the calculation of the period in a spring-masss system
( 1 )
- The equation for the velocity of a simple harmonic motion
( 2 )
where m is the mass of the block, k is the spring constant, A is the amplitude (in this case A = 14 cm) and v is the velocity of the block
Hence
and by reeplacing it in ( 2 ):
In this case for 0.9 s the velocity is zero, that is, the block is in a position with the max displacement from the equilibrium.