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igomit [66]
2 years ago
8

A copper block rests 30.0 cm from the center of a steel turntable. The coefficient of static friction between the block and the

surface is 0.53. The turntable starts from rest and rotates with a constant angular acceleration of 0.50 rad/s2. After what time interval will the block start to slip on the turntable?
pls explain the question briefly
Physics
1 answer:
PIT_PIT [208]2 years ago
3 0

Answer:

refer to the above attachment

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3 years ago
A car traveling 34 mi/h accelerates uniformly for 4 s, covering 615 ft in this time. What was its acceleration? Round your answe
Contact [7]

Answer:

51.94 ft/s²

257.63 ft/s

Explanation:

t = Time taken = 4 s

u = Initial velocity = 34 mi/h

v = Final velocity

s = Displacement = 615 ft

a = Acceleration

Converting velocity to ft/s

34\ mi/h=\frac{34\times 5280}{3600}=49.87\ ft/s

Equation of motion

s=ut+\frac{1}{2}at^2\\\Rightarrow a=2\frac{s-ut}{t^2}\\\Rightarrow a=2\left(\frac{615-49.87\times 4}{4^2}\right)\\\Rightarrow a=51.94\ ft/s^2

Acceleration is 51.94 ft/s²

v=u+at\\\Rightarrow v=49.87+51.94\times 4\\\Rightarrow v=257.63\ ft/s

Final velocity at this time is 257.63 ft/s

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3 years ago
Which action will increase the gravitational force between two objects?
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8 0
3 years ago
Read 2 more answers
A uniform rod of length L is pivoted at L/4 from one end. It is pulled to one side through a very small angle and allowed to osc
ludmilkaskok [199]

Answer:

T= 4.24sec

Explanation:

We are going to use the formula below to calculate.

T=2\pi \sqrt{\frac{L}{g} }

Where T is period

           L is length of rod

       g is acceleration due to gravity =     9.8m/s^{2}

From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this L_{O}

L_{O} = 3/4 * 5.95m

        = 4.4625m

thus   T=2\pi \sqrt{\frac{L_{O} }{g} }

          T=2\pi \sqrt{\frac{4.4625 }{9.8} }

          T= 4.24sec

8 0
4 years ago
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