All categories

2 years ago

Why does gravitational energy increase away from earth

Physics

1 answer:

aliina [53]2 years ago

4 0

Because of the rule of mask.

You might be interested in

Find the gravitational potential energy of a light that has a mass of 13.0 kg and is 4.8 m above the ground

Lynna [10]

Answer:

611.52 <---- Is your answer

Explanation:

If you need explanation mark me as brain list

8 0

2 years ago

If a 2 kg spring is compressed by exerting a force of 10 newtons a distance of 50 cm how much work is done?

Elina [12.6K]

Convert cm into meters,
50/100=0.5m

work done= 10x0.5
=5J

we dont consider the weight of the spring as it acts downwards.

5 0

2 years ago

When reading the printout from a laser printer, you are actually looking at an array of tiny dots.

solniwko [45]

Answer:

The value is $y = 3.097 * 10^{-5} \ m$

Explanation:

From the question we are told that

The diameter of the pupil is $d_p = 4.2 \ mm = 4.2 *10^{-3} \ m$

The distance of the page from the eye $d = 29 \ cm = 0.29 \ m$

The wavelength is $\lambda = 500 \ nm = 500 *10^{-9} \ m$

The refractive index is $n_r = 1.36$

Generally the minimum separation of adjacent dots that can be resolved is mathematically represented as

$y = [ \frac{1.22 * \lambda }{d_p * n_r } ]* d$

$y = [ \frac{1.22 * 500 *10^{-9} }{4.2 *10^{-3} * 1.36} ]* 0.29$

$y = 3.097 * 10^{-5} \ m$

7 0

2 years ago

Which statement is true of the carbon atoms that make up a diamond?

Svetlanka [38]

B, this is because the particles in a solid such as the diamond can not move and even though they are locked into place they still vibrate

8 0

3 years ago

Read 2 more answers

On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole

ohaa [14]

The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
$F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\$
Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation. $\alpha$ is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
$g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\
g_0=g'\\
g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\
ag_0=g_0-w^2r\\
w^2r=g_0(a-1)
$
Our final equation is:
$g=g_0-g_0(a-1)cos^2(\alpha)$
Colatitude is:
$\alpha_c=90^\circ-\alpha$
The answer is:
$g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)$

5 0

2 years ago