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3 years ago

A book weighing 2.0 Newtons is lifted 3.0 meters in 4.0 seconds. How much work was done? SHOW WORK

Physics

1 answer:

qaws [65]3 years ago

7 0

6 Ns is the work done on the book to lift it by 3 m.

Answer:

Explanation:

Work done is the measure of amount of force required to move an object from its initial position to any desired position. So mathematically it can be calculated as the product of force acting on that object to the displacement of that object. Thus, work done on any object or by any object is directly proportional to the force acting on that object and the displacement of that object.

Work done = Force × displacement

As here the force is said to be 2 N and the displacement of the book is said to be 3 m, the work done on the book is

Work done = 2×3 = 6 N.s.

The time has no role to do on the work done calculation to be performed on the object.

Thus, 6 Ns is the work done on the book to lift it by 3 m.

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Select the correct answer from each drop-down menu. A device that uses electricity and magnetism to create motion is called a __

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<h3><u>Answer;</u></h3>

1st drop; Motor

2nd drop; Electricity

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<h3><u>Explanation</u>;</h3>

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Identify two types of simple machines in this compund machine.

sleet_krkn [62]

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Read 2 more answers

The acceleration due to gravity on Earth is 9. 8 m/s2. What is the weight of a 75 kg person on Earth? 9. 8 N 75 N 85 N 735 N.

Illusion [34]

The weight of a 75 kg person on Earth, which has an acceleration due to gravity of 9.8m/s², is 735N.

CALCULATE WEIGHT:

Weight is a force, hence, measured in Newton. It is a function of an object's mass in relation to gravity.

Weight of an object can be calculated using the following formula:

W = mg

Where;

W = weight (N)
m = mass (kg)
g = acceleration due to gravity (m/s²)

The weight of the person in this question can be calculated as follows:

W = 75 × 9.8

W = 735N

Therefore, the weight of a 75 kg person on Earth, which has an acceleration due to gravity of 9.8m/s², is 735N.

Learn more about weight at: brainly.com/question/18554478

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2 years ago

The average power dissipated in a 47 ω resistor is 2.0 w. what is the peak value i 0 of the ac current in the resistor?

pishuonlain [190]

The average dissipated power in a resistor in a ac circuit is:
$P=I_{rms}^2 R$
where R is the resistance, and $I_{rms}$ is the root mean square current, defined as
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where $I_0$ is the peak value of the current. Substituting the second formula into the first one, we find
$P=( \frac{I_0}{\sqrt{2} } )^2 R = \frac{1}{2} I_0^2 R$
and if we re-arrange this formula and use the data of the problem, we can find the value of the peak current I0:
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