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4 years ago

Please help on this one? PLEASE.

Physics

1 answer:

Fantom [35]4 years ago

6 0

Answer: The answer is D

Explanation:

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How many significant figures does 56030 have?<br> 3<br> 4.<br> 5<br> 6

Gre4nikov [31]

The correct answer is 4

8 0

4 years ago

Read 2 more answers

A cart is being pulled forward by a hanging mass. What is the value of the change in its kinetic energy, ΔEk?

Musya8 [376]

The value of change in kinetic energy, ΔEk when a cart is being pulled forward by a hanging mass is Positive

In order words, ΔEk in the above question is positive

<h3>What is kinetic energy?</h3>

Kinetic energy can be defined as the energy a body possess when the body is in motion or due its motion

In conclusion, the value of change in kinetic energy, ΔEk when a cart is being pulled forward by a hanging mass is Positive

Learn more about kinetic energy:

brainly.com/question/25959744

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8 0

2 years ago

A 6.00 kg ball is dropped from a height of 12.0 m above one end of a uniform bar that pivots at its center. The bar has mass 5.0

Andre45 [30]

Answer:

The height of the other ball go after the collision is 2.304 m.

Explanation:

Given that,

Mass of ball = 6.00 kg

Height = 12.0 m

Mass of bar =5.00 kg

Length = 4.00 m

Suppose we need to calculate how high will the other ball go after the collision

We need to calculate the velocity of ball

Using formula of velocity

$v=\sqrt{2gh}$

$v=\sqrt{2\times9.8\times12.0}$

$v=15.33\ m/s$

We need to calculate the angular momentum

Using formula of angular momentum

$l_{before}=mvr$

Put the value into the formula

$l_{before}=6.00\times15.33\times2.0$

$l_{before}=183.96\ kgm^2/s$

We need to calculate the angular momentum

Using formula of angular momentum

$l_{after}=I_{t}\omega$

$l_{after}=(\dfrac{ml^2}{12}+m_{1}r^2+m_{2}r^2)\omega$

Put the value into the formula

$l_{after}=(\dfrac{5\times4.00^2}{12}+6.00\times2.0^2+6.00\times2.0^2)\omega$

$183.96=54.66\omega$

$\omega=\dfrac{183.96}{54.66}$

$\omega=3.36\ rad/sec$

After collision the ball leaves with velocity

We need to calculate the velocity after collision

Using formula of the velocity

$v= r\omega$

$v=2.0\times3.36$

$v=6.72\ m/s$

We need to calculate the height

Using formula of height

$h=\dfrac{v^2}{2g}$

Put the value into the formula

$h=\dfrac{(6.72)^2}{2\times9.8}$

$h=2.304\ m$

Hence, The height of the other ball go after the collision is 2.304 m.

5 0

3 years ago

Can you explain why number 3 is correct?

lukranit [14]

No waves because Q19 waves would going at the surface at regions

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3 years ago

Which of the following has the most biomass?

Crazy boy [7]

Where are the questions so that I can deliver a more accurate answer.

8 0

3 years ago