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2 years ago

Please help with this i am not good with this

Physics

1 answer:

vagabundo [1.1K]2 years ago

6 0

Answer:

Ball 3 (tennis ball)

Explanation:

Formula we are using :

<u>acceleration = Force / mass</u>

<u />

=============================================================

Ball 1 :

⇒ acceleration = 15 N / 0.5 kg

⇒ acceleration = 30 m/s²

===========================================================

Ball 2 :

⇒ acceleration = 15 N / 4.5 kg

⇒ acceleration = 30/9 m/s²

⇒ acceleration = 10/3 m/s²

⇒ acceleration = 3.33 m/s²

===========================================================

Ball 3 :

⇒ acceleration = 15 N/ 0.05 kg

⇒ acceleration = 300 m/s²

============================================================

On comparing, we see that Ball 3 (tennis ball) has the greatest acceleration.

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A mother and her 35.0 -kg child are riding an escalator to the third level of a shopping mall. If the child's gravitational pote

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The increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.

<h3>What is gravitational potential energy?</h3>

The energy that an item has due to its location in a gravitational field is known as gravitational potential energy.

The potential energy increases by 3773 J

PE₂-PE₁=mg(h₂-h₁)

3773 J = 35.0 × 9.81 × (h₂-h₁)

(h₂-h₁) = 10.98

Case 2 ;

ΔPE =?

ΔPE=mg(h₂-h₁)

ΔPE=56.0 × 9.81 ×10.98

ΔPE=6031.97 J.

Hence, the increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.

To learn more about the gravitational potential energy, refer;

brainly.com/question/3884855#SPJ1

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2 years ago

A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

$a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}$

Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

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3 years ago

How can icalculate 3 seconds and 5 meters into a speed?

topjm [15]

To find speed you have to divide distance by time. In this case:

5 meters➗3 seconds = about 1.66666666 and so on m/s.

You could round to 1.67 or 1.7 if you'd like.

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3 years ago

A starship blasts past the earth at 2.0*10^8 m/s .Just after passing the earth, the starship fires a laser beam out its back of

Vilka [71]

Answer:

at the speed of light ( $c=3.0\cdot 10^8 m/s$ )

Explanation:

The second postulate of the theory of the special relativity from Einstein states that:

"The speed of light in free space has the same value c in all inertial frames of reference, where $c=3.0\cdot 10^8 m/s$ "

This means that it doesn't matter if the observer is moving or not relative to the source of ligth: he will always observe light moving at the same speed, c.

In this problem, we have a starship emitting a laser beam (which is an electromagnetic wave, so it travels at the speed of light). The startship is moving relative to the Earth with a speed of 2.0*10^8 m/s: however, this is irrelevant for the exercise, because according to the postulate we mentioned above, an observer on Earth will observe the laser beam approaching Earth with a speed of $c=3.0\cdot 10^8 m/s$ .

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3 years ago

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posledela

The example says skateboard. Something to do a vehicle. Motorcycle or bike are both vehicles which uses the following energy.

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3 years ago

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