PLEEEEASE HELP ME PLEASE WILL GIVE BRAINLIEST

Which of the following statements are true? Check all that apply. Check all that apply. The gravitational force between two obje

AveGali [126]

The gravitational force between two objects is inversely proportional to the square of the distance between the two objects.

The gravitational force between two objects is proportional to the product of the masses of the two objects.

The gravitational force between two objects is proportional to the square of the distance between the two objects. no

The gravitational force between two objects is inversely proportional to the distance between the two objects. no

The gravitational force between two objects is proportional to the distance between the two objects. no

The gravitational force between two objects is inversely proportional to the product of the masses of the two objects. no

3 0

3 years ago

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A bowling ball has a mass of 6 kg. What happens to its momentum when its speed increases from 2 m/s to 4 m/s?

mash [69]

Answer:

B) The initial momentum is 12 kg*m/s, and the final momentum is 24 kg*m/s

Explanation:

The momentum of an object is given by the product between its mass (m) and its velocity (v):

$p=mv$

Let's apply this formula to calculate the initial momentum and final momentum of the ball:

- initial momentum:

$p_i = m v_i = (6 kg)(2 m/s)=12 kg m/s$

- Final momentum:

$p_f = m v_f = (6 kg)(4 m/s)=24 kg m/s$

So, the correct answer is

B) The initial momentum is 12 kg*m/s, and the final momentum is 24 kg*m/s

3 0

3 years ago

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Scientific hypotheses, theories, and laws are all vital to the development of scientific ideas. They are all vital, but they are

Sati [7]

A is correct......

please vote my answer brainliest. thanks!

7 0

4 years ago

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What are the three reasons why nebula contribute more to Stellar formation than any other region in the universe?

monitta

I think it is a, b, and e. Hope it helps! :)

7 0

4 years ago

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At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan

Jet001 [13]

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time $t_1$ , the acceleration vector has direction $\theta_1$ such that

$\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ$

which indicates the particle is situated at a point on the lower left half of the circle, while at time $t_2$ the acceleration has direction $\theta_2$ such that

$\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ$

which indicates the particle lies on the upper left half of the circle.

Notice that $\theta_1-\theta_2=90^\circ$ . That is, the measure of the major arc between the particle's positions at $t_1$ and $t_2$ is 270 degrees, which means that $t_2-t_1$ is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

$\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}$

where $R$ is the radius of the circle and $T$ is the period. We have

$t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s$

and the magnitude of the particle's acceleration toward the center of the circle is

$\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}$

So we find that the path has a radius $R$ of

$7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}$

8 0

3 years ago