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3 years ago

Please help me

Physics

2 answers:

Lisa [10]3 years ago

8 0

Answer:

Explanation:

Your question is quite confusing, particularly the information about microsecond = 0.5.

I'm going to ASSUME that you mean coefficient of static friction μs = 0.5

unfortunately typing a subscript "s" is very difficult and probably leads to such confusion.

I will also ASSUME that the block, and spring, and force vector are all horizontal.

If the force is slowly increased until the block slips, the spring will compress until the force on each end equals the maximum static friction force. As we are not concerned with the compression distance, only the force, we can ignore the spring constant information and simply find the maximum available static friction force.

F = μN

F = μmg

F = 0.5(4.0)(9.8)

F = 19.6 N

Not that it matters, but the spring will have extended or compressed 19.6/0.8 = 24.5 m, which is a very long and very light spring

gtnhenbr [62]3 years ago

4 0

Answer:

Catch these points g thanks

Explanation:

0.1

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Answer:

Explanation:

<u>Gravitational Potential Energy</u>

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3 years ago

A square coil of wire with 15 turns and an area of 0.40 m2 is placed parallel to a magnetic field of 0.75 T. The coil is flipped

Drupady [299]

Answer:

The magnitude of the average induced emf is 90V

Explanation:

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3 years ago

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 8.0 s. At thi

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Answer:

50 revolutions

Explanation:

Data provided:

case I: From rest to top spin

The initial angular speed of the washer, ωi = 0 rev /s

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Time taken, t₁ = 8 s

now,

The angular displacement or the number of revolutions taken (θ₁) is calculated as:

θ₁ = ωi t₁ + (1/2)α₁t₁²

where,

α is the angular acceleration

The angular acceleration can be calculated as:

ωf - ωi = α₁t₁

on substituting the values, we get

8α₁ = 5 - 0

α₁ = 0.625 rev/s²

substituting the values in the equation for the number of revolutions, we get

θ₁ = 0 + (1/2) (0.625)(8)²

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also,

For the case II: From top spin to rest

we have

The initial angular speed, ωi = 5 rev /s

and the final angular speed, ωf = 0 rev /s

Total time taken, t₂ = 12 s

Now, angular acceleration for this case

ωf - ωi = α₂t₂

on substituting the values, we have

12α₂ = 0 - 5

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θ₂ = ωit₂ + (1/2)α₂t₂²

on substituting the values, we have

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θ₂ = 30 rev

Hence,

the total number of revolutions made by the washer during the 20s is

θ = θ₁ + θ₂

θ = 20 rev + 30 rev

θ = 50 revolutions

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natita [175]

Answer:

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