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marysya [2.9K]
2 years ago
12

A block rests on a flat plate that executes vertical simple harmonic motion with a period of 0.74 s. What is the maximum amplitu

de of the motion for which the block does not separate from the plate? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.
Physics
1 answer:
Mumz [18]2 years ago
8 0

Answer:

maximum amplitude  = 0.13 m

Explanation:

Given that

Time period T= 0.74 s

acceleration of gravity g= 10 m/s²

We know that time period of simple harmonic motion given as

T=\dfrac{2\pi}{\omega}

0.74=\dfrac{2\pi}{\omega}

ω = 8.48 rad/s

ω=angular frequency

Lets take amplitude = A

The maximum acceleration given as

a= ω² A

The maximum acceleration should be equal to g ,then block does not separate

a= ω² A

10= 8.48² A

A=0.13 m

maximum amplitude  = 0.13 m

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Centripetal Force (Fcp) = ?

His arm length = Radius (R) = 0.75 m

Discus velocity = Linear Velocity (V) = 5 m/s

Discus mass (m) = 2 kg

Centripetal Acceleration (Acp) = V^2/R or W^2 x R
In this case i will use the V^2/R formula, because it uses the discus velocity (V).

fcp = m \times acp \\ fcp = m \times {v}^{2} \div r \\ fcp = 2 \times {5}^{2} \div 0.75 \\ fcp = 2 \times 25 \div 0.75 \\ fcp = 50 \div 0.75
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Answer: Last option, 66 N.
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Decreasing the distance between Hox and Blox, increasing the mass of Hox, or increasing the mass of Hox and Blox.

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The gravity force is directly proportional to the mass of the bodies and inversely proportional to the square of the distance that separates them.

Or

If we decrease the distance between both planets (Hox and Blox), the gravitational pull between them will increase.  

On the other hand, if we keep the distance between Hox and Blox, but we increase the mass of one of them, or increase the mass of both, the gravitational pull between them will also increase.

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A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnet
Mumz [18]
<h3>Question:</h3>

A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.

<h3>Answer:</h3>

1.6nT [in the negative z direction]

<h2>Explanation:</h2>

The magnetic field, B, due to a distance of finite value b, is given by;

B = (μ₀IL) / (4πb\sqrt{b^2 + L^2})                -----------(i)

Where;

I = current on the wire

L = length of the wire

μ₀ = magnetic constant = 4π × 10⁻⁷ H/m

From the question,

I = 20A

L = 2.0cm = 0.02m

b = 5.0m

Substitute the necessary values into equation (i)

B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0 \sqrt{5.0^2 + 0.02^2})

B = (10⁻⁷ x 20 x 0.02) / (5.0 \sqrt{5.0^2 + 0.02^2})

B = (10⁻⁷ x 20 x 0.02) / (5.0 \sqrt{25.0004})

B = (10⁻⁷ x 20 x 0.02) / (25.0)

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B = 1.6nT

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Answer:

F= 2666.66 N

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