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Sunny_sXe [5.5K]
3 years ago
5

Why is the work output of a machine never equal to the work input? A. Some work input is used to overcome friction. B. Input dis

tance never equals output distance. C. Some work output is used to overcome friction.
Physics
2 answers:
SVETLANKA909090 [29]3 years ago
6 0
The answer is A. <span>Some work input is used to overcome friction. </span>
Reptile [31]3 years ago
3 0

Answer:

A. Some work input is used to overcome friction.

Explanation:

As we know that in real conditions it is not possible to form a surface which do not have any friction or we can say that it is not possible to have a surface which is 100% smooth

So here when a machine absorbs some energy as its input work then some part of this energy is used against frictional force between its moving part and hence the output energy is less than the energy that it absorb.

So output energy is never same as the input energy of a machine.

So correct answer would be

A. Some work input is used to overcome friction.

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Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

T-mg = m\frac{v^2}{R}

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

m \frac{v^2}{R} is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

T=m\frac{v^2}{R}

And by substituting the numerical values, we find

T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

T+mg=m\frac{v^2}{R}

And substituting the numerical values and re-arranging it, we find

T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

T=0\\mg = m\frac{v^2}{R}

and re-arranging the equation, we find

v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s

7 0
3 years ago
A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s
gulaghasi [49]

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T=2\cdot 2.5 s=5.0 s

The frequency of the waves is the reciprocal of the period:

f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz

The wavelength instead is just the distance between two consecutive crests, so

\lambda=4.8 m

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v=\lambda f=(4.8 m)(0.20 Hz)=0.96 m/s

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The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

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In this case, the amplitude of the wave would be lower. In fact,

d = 0.30 m

So the amplitude would be

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How do you break apart a
Tcecarenko [31]

Answer:

B

Explanation:

Depends Mostly on bonds electrolysis can be used, chemical bonding like additional of water or by heating back to their elements.

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2 years ago
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Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. The energies of atoms a
lawyer [7]

Answer:

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A ball of mass m, attached to the end of a horizontal cord, is rotated in a circle of radius r on a frictionless horizontal surf
kirza4 [7]

Answer:v=\sqrt{\frac{FL}{m}}

Explanation:

Given

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3 years ago
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