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3 years ago

14

A construction worker uses 1,000 J of work to screw two boards together. The screw does 855 J of work. Calculate the efficiency

of the screw.
PLZ HELP!!!

1 answer:

Dovator [93]3 years ago

7 0

Answer:

The efficiency of the screw is 85.5%

Explanation:

to calculate efficiency you have to do

efficiency=energy output/ energy input × 100%

so e=855/1000 × 100%=85.5%

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A ball is thrown straight up into the air and passes a window 3 seconds after being released. It passes the same window on its w

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The initial velocity of the ball is 55.125 m/s.

<h3>Initial velocity of the ball</h3>

The initial velocity of the ball is calculated as follows;

During upward motion

h = vi - ¹/₂gt²

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h = vi - 44.1 ----------------- (1)

During downward motion

h = vi + ¹/₂gt²

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Thus, the initial velocity of the ball is 55.125 m/s.

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1 year ago

According to Newton’s first law of motion, what will an object in motion do when no external force acts on it?

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By definition, we have to:

Newton's first law states that any object will remain in a state of rest or with a uniform rectilinear motion unless an external force acts on it.

Therefore, according to the first law of Newton, if the object is already in motion and has no force acting on it then, it will remain with a uniform rectilinear motion.

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4 years ago

Read 2 more answers

A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.

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Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, $KE_{0} = 0.5 mu^{2}$

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, $KE_{f} = 0.5mv^{2}$

a) Using the principle of energy conservation,

$KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s$

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

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