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3 years ago

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Calculate the efficiency for each machine, then answer the question. Machine A has an input work of 1,000 J and an output work o

f 250 J. Machine B has an input work of 500 J and an output work of 350 J. Machine C has an input work of 200 J and an output work of 150 J. Which machine is the most efficient?

2 answers:

Usimov [2.4K]3 years ago

7 0

Machine C. Machine A has an efficiency of 250/1000 = 0.25 = 25%, Machine B's efficiency is 350/500 = 0.7 = 70% and Machine C has an efficiency of 150/200 = 0.75 = 75%.

kati45 [8]3 years ago

7 0

Explanation:

Efficiency is defined as the measure of amount of work done by a system.

Mathematically, $Efficiency = \frac{Energy output}{Energy input} \times 100$

For machine A efficiency will be as follows.

Efficiency = $\frac{Energy output}{Energy input} \times 100$

= $\frac{250 J}{1,000 J} \times 100$

= 25%

For machine B efficiency will be as follows.

Efficiency = $\frac{Energy output}{Energy input} \times 100$

= $\frac{350 J}{500 J} \times 100$

= 70%

For machine C efficiency will be as follows.

Efficiency = $\frac{Energy output}{Energy input} \times 100$

= $\frac{150 J}{200 J} \times 100$

= 75%

Thus, we can see that efficiency is maximum in machine C, therefore, machine C is the most efficient.

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What is the energy of a rock with a mass of 10.2 kg on a cliff that is 300 m height?

Anuta_ua [19.1K]

The potential energy of the rock is 30,000 J

Explanation:

The mechanical energy of an object is equal to the sum of its gravitational potential energy (PE) and its kinetic energy (KE):

$E=PE+KE$

where

PE is the gravitational potential energy, which is the energy possessed by the object due to its position in the gravitational field

KE is the kinetic energy, which is the energy possessed by the object due to its motion

In this problem, the rock is at rest, so its kinetic energy is zero:

KE = 0

Therefore, the energy of the rock is just equal to its potential energy, which is:

$E=PE=mgh$

where

m = 10.2 kg is the mass of the rock

$g=9.8 m/s^2$ is the acceleration of gravity

h = 300 m is the height of the rock above the ground

Substituting and solving, we find

$PE=(10.2)(9.8)(300)=30,000 J$

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3 years ago

A meter stick is free to rotate about an axis through one of its end. Find the force F needed to balance this meter stick if the

Mumz [18]

Answer:

Explanation:

Component of force perpendicular to stick

= F Sin 60°

=√3 / 2 F.

Taking torque about the other end

= √3 / 2 F x 1 Nm

Weight of stick = 60 gm

= 60 x 10⁻³ kg

= 60 x 10⁻³ x 9.8 N

= .588 N

This weight will act from the middle point of stick so torque about the

other end

= .588 x 1 Nm

Balancing these two torques we have

.588 = √3 /2 F

$F=\frac{2\times0.588}{\sqrt{3} }$

F = 0.679 N

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3 years ago

(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit

Ainat [17]

Answer:

a

The number of fringe is z = 3 fringes

b

The ratio is $I = 0.2545I_o$

Explanation:

a

From the question we are told that

The wavelength is $\lambda = 600 nm$

The distance between the slit is $d = 0.117mm = 0.117 *10^{-3} m$

The width of the slit is $a = 35.7 \mu m = 35.7 *10^{-6}m$

let z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is and this mathematically represented as

$z = \frac{d}{a}$

Substituting values

$z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}$

z = 3 fringes

b

From the question we are told that the order of the bright fringe is n = 3

Generally the intensity of a pattern is mathematically represented as

$I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]$

Where $I_o$ is the intensity of the central fringe

And Generally $sin \theta = \frac{n \lambda }{d}$

$I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]$

$I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]$

$I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]$

$I = I_o (1)(0.2545)$

$I = 0.2545I_o$

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3 years ago

A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it

Savatey [412]

The charge of the object must be $1.11 \times e^{-5} \text { coulomb }$

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

$Electric field, E=\frac{\text { Force }(F)}{q}$

Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

By substituting the given values, we get

$q=\frac{F}{E}=\frac{0.05 N}{4500 \mathrm{N} / \mathrm{c}}=1.11 \times e^{-5} \text { coulomb }$

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3 years ago

Which object has more gravitational potential energy? Use PE = m × g × h, where g = 9.8 meters/second2.

Andrew [12]

Gravitational potential energy can be calculated using the formula <span>PE = m × g × h, where g is the gravitational acceleration and is constant hence the energy is dependent directly to mass and the height of the object. Hence more PE is registered when the object is heavier and/or at greater initial height. </span>

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3 years ago

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