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3 years ago

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Besides a reduction in friction, the only way to increase the amount of work output of a machine is to _____ the work input. Dec

rease reduce increase.

1 answer:

Vedmedyk [2.9K]3 years ago

7 0

Answer:

besides a reduction in friction, the only way to increase the amount of work output of a machine is to Increase the work input

i

Explanation:

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Calculate the average orbital speed of Ceres in

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3 years ago

A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr

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The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
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$f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz$
and the angular frequency is
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In a spring-mass system, the maximum velocity of the object is given by
$v_{max} = A \omega$
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
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4 years ago

A ship sets out to sail to a point 120 km due north. an unexpected storm blows the ship to a point 100 km due east of its starti

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If you draw the problem, it would look like that shown in the attached picture. The total length the ship will now travel can be solved using the Pythagorean theorem. The solution is as follows:

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4 years ago

If the weight of the ruler is one Newton ,Gc cannot have a value more than 25cm

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If the length of the ruler is 50 cm, the center of gravity cannot be greater than 25 cm.

The given parameters:

Weight of the ruler = 1 N

<h3>What is center of gravity (CG)?</h3>

Center of gravity is the point at which the weight of an object is concentrated.

Let the length of the ruler = L

The center of the gravity of the ruler is calculated as follows;

$X_{CG} = \frac{W(L_0) + W(L -X_{CG})}{W} \\\\X_{CG} = \frac{1(0) + 1(L -X_{CG})}{1}\\\\X_{CG} = L - X_{CG}\\\\X_{CG } + X_{CG} = L\\\\2X_{CG} = L\\\\X_{CG} = \frac{L}{2} \\\\when , \ L = 50 \ cm\\\\X_{CG} = \frac{50}{2} \\\\X_{CG} = 25 \ cm$

Thus, if the length of the ruler is 50 cm, the center of gravity cannot be greater than 25 cm. This may change if the length of the ruler changes because the center of gravity of uniform ruler depends on the length of the ruler.

Learn more about center of gravity here: brainly.com/question/6765179

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