A disk of radius 10 cm is pulled along a frictionless surface with a force of 16 N by a string wrapped around the edge. At the i
1 answer:
Answer:
t = 0.2845Nm (rounded to 4 decimal places)
Explanation:
The disk rotates at a distance of an arc length of 28cm
Arc length = radius × central angle × π/180
28cm = 10cm × central angle × π/180
Central angle = × 180/π ≈ 160.4°
Torque (t) = rFsin(central angle) , where F is the applied force
Radius in meters = 10/100 = 0.1m
t = 0.1m × 16N × sin160.4°
t = 0.2845Nm (rounded to 4 decimal places)
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is the magnitude of the acceleration
In the horizontal direction.
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(1)
where T is the tension in the string.
We also know that the bob is in equilibrium along the vertical direction: so the vertical component of the tension must be equal to the weight of the bob,
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Dividing eq.(1) by eq(2), we get:
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Answer:
D. 550 Hz
Explanation:
Frequency for string is given by
where v is the speed of the wave, L = m is the length
Frequency is inversely proportional to the Length of the string
from above relation we can write
The newly shortened string has 4/5 the length of the full string .i,e
Hence option D is correct