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dexar [7]
3 years ago
6

A force of 45 newtons is applied on an object, moving it 12 meters away in the same direction as the force. What is the magnitud

e of work done on the object by this force? Part A: Enter the variable symbol for the quantity you need to find. Use your keyboard and the keypad to enter your answer. Then click Done.
Physics
1 answer:
NARA [144]3 years ago
6 0
<h2>Answer: 540 J</h2>

Explanation:

The Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.  

Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

W=(F)(d) (1)  

In this case both (the force and the distance in the path) are parallel (this means they are in the same direction), so the work W performed is the product of the force exerted to push the box F=45N by the distance traveled d=12m.

Hence:  

W=(45N)(12m)   (2)

W=540Nm=540J

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The solution to the questions are given as

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<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}

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\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}

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c)

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