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3 years ago

An object has a mass of 7 kg and is accelerating at 4 m/s2. How far would the object move if it took 168 J of work to move it?

Physics

1 answer:

KIM [24]3 years ago

6 0

Answer:

Explanation:

use work =force×distance

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Which choice correctly ranks these items from smallest to largest?

Vera_Pavlovna [14]

Answer:

your answer is: electron → carbon atom → quantum dot → E. coli bacteria cell → comma

Explanation:

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3 years ago

Darcy is going to make raspberry jam for the county fair.

valentinak56 [21]

Answer:

She can make have 30 jars with raspberries in them with 50 left over.

Explanation:

1,700 divided by 55

30 equally

but 50 left over

This means that she can make have 30 jars with raspberries in them with 50 left over.

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3 years ago

If runner A is running at 7.50 m/s and runner B is running at 7.90 m/s, how long will it take runner B to catch runner A if runn

yaroslaw [1]

Answer:

t= 137.5 s

Explanation:

So if we are wanting to figure out how long it takes runner B to catch runner A. we must first set the slope of each runner equal to one another

<u>Slopes:</u>

Runner A: y = 7.50x + 55

Runner B: y = 7.90 x

sooooo

7.50 x + 55 = 7.90 x

- 7.50 x - 7.50 x

55 = .40 x

55/.40 = .40 x / .40

x = 137.5 s

t= 137.5 s

7.50 * 137.5 + 55 =1086.25 m

7.90 * 137.5 = 1086.25 m

3 0

2 years ago

A 2.5-kg ball and a 5.0-kg ball have an elastic collision. Before the collision, the 2.5-kg ball was at rest and the other ball

lesantik [10]

The kinetic energy of 2.5 kg ball after collision is 27.09 J.

Answer:

Explanation:

In elastic collision, the sum of momentum of the objects before collision will be equal to the sum of momentum of the objects after collision.

We know that momentum is the product of mass and velocity acting on any object.

So, the conservation of energy in elastic collision leads to following equation:

$M_{1} u_{1} +M_{2} u_{2}=M_{1} v_{1}+M_{2} v_{2}$

Since, the momentum is conserved ,the kinetic energy will also be conserved in elastic collision. So

$M_{1} u_{1} ^{2}+M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+ M_{2}v_{2} ^{2}$

Since initial velocity for M1 ball is zero, then

$M_{2} u_{2}=M_{1} v_{1}+M_{2} v_{2}$

and

$M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+ M_{2}v_{2} ^{2}$

So, on solving all the above equation, we get an equation for velocity and that is

$\frac{2M_{2}u_{2} }{(M_{1}+M_{2} }$ =final velocity of ball with mass 2.5 kg

$v = \frac{2(5*3.5)}{2.5+5}=4.67 m/s$

So kinetic energy will be 1/2 mv2

Kinetic energy of 2.5 kg ball is $\frac{1}{2}*2.5*(4.67)^{2} =27.09 J$

So the kinetic energy of 2.5 kg ball after collision is 27.09 J.

6 0

3 years ago

Acceleration is rate of change of <br>A-Position<br>B-Time<br>C-Velocity<br>D-Speed

Drupady [299]

<h3>Answer:</h3>

[C] Velocity.

<h3>Explanation:</h3>

<u>As we know that</u>,

a = v - u/t

<u>where, a = acceleration, v = final velocity, u = initial velocity and t = time taken to travel</u>.

8 0

3 years ago

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An object has a mass of 7 kg and is accelerating at 4 m/s2. How far would the object move if it took 168 J of work to move it?​

An object has a mass of 7 kg and is accelerating at 4 m/s2. How far would the object move if it took 168 J of work to move it?