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2 years ago

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A rope is being pulled with a force of 366 N parallel to the horizontal surface right, five men are pulling the same rope parall

el to the horizontal surface but toward left. what amount of force should an individual apply towards left to keep the rope in equilibrium (suppose that each of five men is contributing same force)

1 answer:

lakkis [162]2 years ago

3 0

The amount of force applied by individual man to balance the force at the right is 73.2 N.

<h3>Amount of force applied by individual man</h3>

The amount of force applied by individual man to balance the force at the right is calculated as follows;

$F_{right} = F_{left}\\\\F_{left} = F_1 + F_2 + F_3 + F_4 + F_5\\\\F _1 = F_2 = F_3 = F_4 = F_5\\\\F_1 = \frac{F_{right}}{5} = \frac{366}{5} = 73.2 \ N$

Thus, the amount of force applied by individual man to balance the force at the right is 73.2 N.

Learn more about forces here: brainly.com/question/12970081

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3 years ago

A rope is wrapped around a pulley many times. The pulley can be modeled as a solid disk of radius R and mass M, and a mass mA ha

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The magnitude of the tangential acceleration of the hanging mass is 2mg/MR

<h3>Tangential acceleration of the hanging mass</h3>

The tangential acceleration of the hanging mass around the pulley is determined from the principle of conservation of angular momentum as shown below;

τ = Iα

Where;

I is the moment of inertia
α is the angular velocity

$\alpha = \frac{\tau}{I} \\\\\alpha = \frac{mgR}{3/2MR^2} \\\\\alpha = \frac{2mgR}{3MR^2} \\\\\alpha = \frac{2mg}{3MR}$

Where;

m is the hanging mass
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The tangential acceleration is calculated as follows;

$a = \alpha R\\\\a = \frac{2mg}{3MR} \times R\\\\a = \frac{2mg}{3M}$

Thus, the magnitude of the tangential acceleration of the hanging mass is 2mg/MR

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3 years ago

A parallel-plate capacitor has plates with an area of 451 cm2 and an air-filled gap between the plates that is 2.51 mm thick. Th

Nostrana [21]

To solve this problem we will apply the concepts related to Energy defined in the capacitors, as well as the capacitance and load. From these three definitions we will build the solution to the problem by defending the energy with the initial conditions, the energy under new conditions and finally the change in the work done to move from one point to the other.

Energy in a capacitor can be defined as

$E = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C}$

Here,

V = Potential difference across the capacitor plates

Q = Charge stored on the capacitor plates

At the same time capacitance can be defined as,

$C = \epsilon_0 (\frac{A}{d})$

Here,

$\epsilon_0 =$ Vacuum permittivity constant

A = Area

d = Distance

Replacing with our values we have that,

$C = (8.85*10^{-12})(\frac{0.0451}{2.51*10^{-3}})$

$C = 1.5901*10^{-10}F$

PART A) Energy stored in the capacitor is

$E = \frac{1}{2} CV^2$

$E = \frac{1}{2} (1.5901*10^{-10})(575)^2$

$E = 2.628*10^{-5}J$

PART B) We know first that everything that the load can be defined as the product between voltage and capacitance, therefore

$Q = CV$

$Q = (1.59*10^{-10})(575)$

$Q = 9.1425*10^{-8}C$

Now if $d = 10.04*10^{-3}m$ we have that the capacitance is

$C = \epsilon_0 (\frac{A}{d})$

$C = (8.85*10^{-12})(\frac{0.0451}{10.04*10^{-3}})$

$C = 3.9754*10^{-11}F$

Then the energy stored is

$E = \frac{1}{2} \frac{Q^2}{C}$

$E = \frac{1}{2} (\frac{(9.1425*10^{-8})^2}{3.9754*10^{-11}})$

$E = 1.051*10^{-4} J$

PART C) The amount of work or energy required to carry out this process is the difference between the energies obtained, therefore

$W = 1.051*10^{-4} J -2.628*10^{-5}J$

$W = 7.882*10^{-5} J$

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3 years ago

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alukav5142 [94]

Answer:

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Explanation:

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3 years ago

Read 2 more answers