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3 years ago

Could you pick up a nail using the curved part of the horseshoe magnet farthest from the poles?

Physics

1 answer:

nikitadnepr [17]3 years ago

5 0

Yes

Explanation:

One can pick up a nail using the curved part of the horseshoe magnet farthest from the poles.

A horse shoe magnet has magnetic fields all around it.

In the presence of magnetic force fields, any magnetic object will be attracted to it.
This is the case with any magnet.
A magnet is any object with magnetic fields all around it.
This field causes attraction to any magnetic objects especially metals.

learn more:

Electromagnet brainly.com/question/2191993

#learnwithBrainly

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3 years ago

jill's car has a maximum acceleration of 8.7 miles per hour per second. how many seconds does it take her to acceleration from 0

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T = ?
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3 years ago

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I would like help with this physics problem

Darina [25.2K]

(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.

Recall that

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We have $v_f=-v_i$ , so that

$-2v_i=at\implies-2\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)=\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\implies t=8\,\mathrm s$

(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.

(d) The position of the ball $x_f$ at time $t$ is given by

$x_f=x_i+v_it+\dfrac12at^2$

Take the starting position to be the origin, $x_i=0$ . Then after 6 seconds,

$x_f=\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2=42\,\mathrm m$

so the ball is 42 m away from where it started.

We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:

$x_f-x_i=\dfrac{v_i+v_f}2t\implies42\,\mathrm m=\dfrac{8\,\frac{\mathrm m}{\mathrm s}+v_f}2(6\,\mathrm s)\implies v_f=6\,\dfrac{\mathrm m}{\mathrm s}$

Since the velocity is positive, the ball is still moving up the incline.

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3 years ago