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3 years ago

What is the magnitude of the net torque on the pulley about the axle?

Physics

1 answer:

jeyben [28]3 years ago

6 0

Answer: Torque= tangential force × radius of the pulley

Explanation:

Pulley is a simple machine which helps us to lift loads by applying the force in a convenient direction. It consists of a grooved wheel which is free to rotate about an axle passing through the center of the pulley.

For a pulley, the torque $T=$ tangential force × radius of the pulley

Pulleys can be used in combination also to achieve a desired output.

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photoshop1234 [79]

Answer:

she'd get thrown forward

Explanation:

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3 years ago

Technician A says that you can usually depressurize a brake accumulator by turning the ignition switch ""off"" to disable the el

Rzqust [24]

Answer:

Both Technician A and Technician B

Explanation:

In order to gain a better understanding of the solution above let define some terms

Break Accumulator

We can define a break accumulator as storage that that helps generate the required pressure in order for the breaking system to respond faster this accumulator is charged by turning the steering wheel slowly at once from lock to lock now this build the pressure in the accumulator and one way to depressurize is it is by turning the ignition switch ""off""

Now a scan tool is a device that can interface with a car it can also be used to diagnose a car an get the diagnostic information to help in the cars diagnoses and also be used to reprogram a car

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3 years ago

A simple pendulum consists of a brass sphere of mass m = 1 kg suspended on a string of length L ≈ 1 meter. The pendulum oscillat

hram777 [196]

Answer:

2. Using a shorter string of length L ′ ≈ 0.25 meters

5. Using a shorter string of length L ′ ≈ 0.5 meters

Explanation:

The period of a pendulum is given by

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

where

L is the length of the pendulum

g is the acceleration due to gravity

We see from the formula that the period of the pendulum depends only on its length, not on its mass or its amplitude of ocillation. Therefore, the only alterations that can change the period of the pendulum are the ones where its length is changed.

Moreover, we notice that the period is proportional to the square of the length: this means that in order to decrease the period of the pendulum (the problem asks us which alterations will reduce the period of the pendulum from 2 s to 1 s), the length of the pendulum should also be reduced.

Therefore, the only alterations that will reduce the period of the pendulum are:

2. Using a shorter string of length L ′ ≈ 0.25 meters

5. Using a shorter string of length L ′ ≈ 0.5 meters

6 0

3 years ago

A plastic sphere floats in water with 50.0% of its volume submerged. This same sphere floats in glycerin with 40.0% of its volum

Natasha_Volkova [10]

Explanation:

weight of water displaced=weight of sphere,

Since, plastic sphere floats in water with 50.0% of its volume submerged

so, water=2 sphere,

so sphere's density = 1/2 of water's density

Now, weight of glycerin displaced=weight of sphere,

Also given This same sphere floats in glycerin with 40.0% of its volume submerged.

so glycerin =2.5 sphere,

so sphere's density = .4 of glycerin's = 1/2 of water's.

So glycerin = 1.25 water

6 0

3 years ago

A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 8.90 kg box that is sitting on the horizontal,

Lady_Fox [76]

Answer:

6.875 m/s

Explanation:

The force is variable which is given by

F(x) = 18 - 0.53 x

mass of the box, m = 8.9 kg

initially it is at rest at x = 0

Let the velocity is v after travelling a distance of 15 m.

According to the work energy theorem, the work done by all the forces is equal to the change in kinetic energy of the body.

Work done = change in kinetic energy

$\int \overrightarrow{F}.d\overrightarrow{x}=\Delta K.E$

$\int_{0}^{15} \left ( 18-0.53 x \right )dx=\frac{1}{2}\times m \left ( v^{2}-u^{2} \right )$

$\left ( 18x-0.265x^{2} \right )_{0}^{15}=\frac{1}{2}\times 8.9\times \left ( v^{2}-0^{2} \right )$

18 x 15 - 0.265 x 15 x 15 = 4.45 x v²

270 - 59.625 = 4.45 v²

v² = 47.275

v = 6.875 m/s

Thus, the final velocity of the box is 6.875 m/s.

4 0

3 years ago