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3 years ago

An 800 N box is pushed across a level floor for a distance of 5.0 m with a force of 400 N. How much work was done on this box.

Physics

1 answer:

AveGali [126]3 years ago

4 0

Answer: 2000 J.

Explanation: Since work is force*displacement, we just have to multiply the force by the distance: w = f*d = 400 N*5.0 m = 2000 J.

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No the gravity of the moon pulls the water making high tide

Explanation:

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3 years ago

The distance from one crest to the next is the ________.

nikitadnepr [17]

Answer:

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Explanation:

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3 years ago

John and Caroline go out for a walk one day. This graph represents their distance from home.

Serga [27]

Answer:

Option C is the correct answer.

Explanation:

We have velocity of a body = Change in position/ Time.

Considering first portion of graph,

Change in position = 30 - 0 = 30 m

Time = 0.75 hours = 45 minutes = 2700 seconds

Velocity = 30/2700 = 0.011 m/s

Considering second portion of graph,

Change in position = 30 - 30 = 0 m

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Considering third portion of graph,

Change in position = 0 - 30 = -30 m

Time = 0.75 hours = 45 minutes = 2700 seconds

Velocity = -30/2700 = -0.011 m/s

So firstly they walked in one direction(positive direction), then they were still(velocity is zero), then they walked in the opposite direction( velocity is negative).

Option C is the correct answer.

3 0

3 years ago

Read 2 more answers

The Celsius temperature scale is based on which of the following

Lapatulllka [165]

Freezing.................

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3 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

3 years ago