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Artemon [7]
3 years ago
10

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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A Carnot engine's operating temperatures are 240 ∘C and 20 ∘C. The engine's power output is 910 W . Part A Calculate the rate of
scoray [572]

Answer:1200

Explanation:

Given data

Upper Temprature\left ( T_H\right )=240^{\circ}\approx 513

Lower Temprature \left ( T_L\right )=20^{\circ}\approx 293

Engine power ouput\left ( W\right )=910 W

Efficiency of carnot cycle is given by

\eta =1-\frac{T_L}{T_H}

\eta =\frac{W_s}{Q_s}

1-\frac{293}{513}=\frac{910}{Q_s}

Q_s=2121.954 W

Q_r=1211.954 W

rounding off to two significant figures

Q_r=1200 W

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3 years ago
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Answer:

453 gm

Explanation:

<u>Immersed </u>objects are buoyed up by force equal to mass of displaced liquid

400 + 53 = 453 gm  in air

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1 year ago
What is 60mph (miles per hour) in meters per second? ( A mile is 5280ft)<br> please someone help me
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3 years ago
A kayakeris paddling 2.50 m/s at an angle of 45° (northeast) and the current is moving 1.25 m/s at an angle of 315° (southeast).
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The kayaker has velocity vector

<em>v</em> = (2.50 m/s) (cos(45º) <em>i</em> + sin(45º) <em>j</em> )

<em>v</em> ≈ (1.77 m/s) (<em>i</em> + <em>j</em> )

and the current has velocity vector

<em>w</em> = (1.25 m/s) (cos(315º) <em>i</em> + sin(315º) <em>j</em> )

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The kayaker's total velocity is the sum of these:

<em>v</em> + <em>w</em> ≈ (2.65 m/s) <em>i</em> + (0.884 m/s) <em>j</em>

That is, the kayaker has a velocity of about ||<em>v</em> + <em>w</em>|| ≈ 2.80 m/s in a direction <em>θ</em> such that

tan(<em>θ</em>) = (0.884 m/s) / (2.65 m/s)   →   <em>θ</em> ≈ 18.4º

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