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Greeley [361]
3 years ago
5

A 1.10-kg wrench is acting on a nut trying to turn it. The length of the wrench lies directly to the east of the nut. A force 15

0.0 N acts on the wrench at a position 15.0 cm from the center of the nut in a direction 30.0° north of east. What is the magnitude of the torque about the center of the nut
Physics
1 answer:
gtnhenbr [62]3 years ago
7 0

Answer:

τ = 11.25 Nm

Explanation:

Given,

Mass of wrench, m = 1.10 Kg

Force on the wrench, F = 150 N

position of force, r = 15 cm = 0.15 m

angle of nut = 30°

We know,

torque = Force x distance

F = 150 sin 30°

F = 75 N

τ = 75 x r

τ = 75 x 0.15

τ = 11.25 Nm

Torque at the center of the nut is equal to 11.25 N.

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Answer:

The correct answer to the question is

B. It always decreases

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To solve the question, we note that the foce of gravity is given by

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G= Gravitational constant

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m₂ = mass of second object

r = the distance between both objects

If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have

F_{G2} =\frac{Gm_1(2m_2)}{(2r)^2} = \frac{2}{4} \frac{Gm_1m_2}{r^2}

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The deliberate radiation of electromagnetic (EM) energy to degrade or neutralize the radio frequency long-haul supervisory contr
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Best explains Jamming

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<em>The deliberate radiation of electromagnetic (EM) energy to degrade or neutralize the radio frequency long-haul supervisory control and data acquisition (SCADA) communications links, best explains what?</em>

Jamming is defined as the blocking or interference with authorized wireless communications. it's a problem  in personal area network wireless technologies. Jamming can occur inadvertently due to high levels of noise .

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In what sense is energy from coal actually solar energy
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A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave
Oliga [24]

Answer:3.51

Explanation:

Given

Coefficient of Friction \mu =0.4

Consider a small element at an angle \theta having an angle of d\theta

Normal Force=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}

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Friction f=\mu \times Normal\ Reaction

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and T+dT-T=f=\mu Td\theta

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\frac{T_2}{T_1}=e^{\mu \pi}

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\frac{T_2}{T_1}==e^{1.256}

\frac{T_2}{T_1}=3.51

7 0
3 years ago
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