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pickupchik [31]
3 years ago
15

300 Jules energy are used to push an object with a force of 75N. what is the maximum distance the object can move?

Physics
1 answer:
Artyom0805 [142]3 years ago
6 0

Energy consumed in doing the work = 300 Joules

Force applied on the object = 75 N

Let the distance moved by the object be d.

Work done by the force is determined by the force applied and the displacement happened in the direction of the force applied.

Work done = Force x displacement

300 = 75 x d

d = \frac{300}{75}

d = 4 m

Hence, the maximum distance moved by the object = 4 meters

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Answer:

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A 2kg blob of putty moving at 4 m/s slams into a 6kg blob of putty at rest what is the speed of the to stick together blobs imme
andrew11 [14]
<h2>Answer</h2>

1m/s

<h2>Explanation</h2>

Given that:

<em>Mass of first blob = 2kg = m1</em>

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<em>Mass of second blob = 6kg = m2</em>

<em>Velocity of blob = 0m/s = v2</em>

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To find:

<em>Final velocity = Vf</em>

<em />

<em>This question is of inelastic collision which is any collision between objects in which some energy is lost.</em>

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<h3>Formula to be use:</h3><h2>(m1*v1) + (m2*V2) = Vf(m1 + m2)</h2>

(2*4) + (6*0) = Vf(2+6)

8 + 0 = Vf(8)

8 = Vf(8)

Vf = 1 m/s

So the speed of two blobs immediately after colliding = 1 m/s

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What is the definition for model system
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If you wanted to know how fast a tiger can run, which measurements would you use?
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A stone is tied to a string and whirled at constant speed in a horizontal circle. The speed is then doubled without changing the
Tanya [424]

Answer:

Afterward the magnitude of the acceleration of the stone is four times as great

Explanation:

A stone tied to a string and whirled at a constant speed in a horizontal circle will have a centripetal acceleration given by

a = \frac{v^{2} }{r}

Where a is the centripetal acceleration

v is the speed

and r is the radius of the circle

Here, the radius of the circle is the length of the string.

Now, from the question

The speed is then doubled without changing the length of the string,

Let the new speed be v_{2}, that is

v_{2} = 2v

and without changing the length of the string means radius r is constant.

To determine the magnitude of the acceleration of the stone afterwards,

Let the new acceleration be a_{2}.

Then we can write that

a_{2} = \frac{v_{2}^{2}  }{r}

From

a = \frac{v^{2} }{r}

v = \sqrt{ar}

Recall that

v_{2} = 2v

∴ v_{2} = 2\sqrt{ar}

Now, we will put the value of v_{2} into

a_{2} = \frac{v_{2}^{2}  }{r}

Then,

a_{2} = \frac{(2\sqrt{ar}) ^{2}  }{r}

a_{2} = \frac{4ar }{r}

a_{2} = 4a

The new magnitude of the acceleration of the stone is four times the initial acceleration.

Hence,

Afterward the magnitude of the acceleration of the stone is four times as great

3 0
3 years ago
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