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Oxana [17]
3 years ago
6

_________is an object which is in orbit around another celestial body in space. It can be natural or artificial.

Physics
1 answer:
Delvig [45]3 years ago
4 0

i think its natural satelites

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oee [108]
I think is limestone
7 0
3 years ago
Need help as fast as possible please
DaniilM [7]
Regular reflections occurs when light reflects off a very smooth surface and forms a clear image.
Diffuse reflection: occurs when light reflects off a rough surface and forms a blurry image or no image at all.

Similarities: they both hit the surface, but the difference is one of them give clear image the other one gives blurry image or no image at all.

Hope this helped!
7 0
3 years ago
Someone help meeeee! please
Ratling [72]

Answer:

This looks like an example of combustion

Explanation:

5 0
3 years ago
A projectile is fired upward with an initial speed vo on an airless world. A short time later, it comes back down and has a fina
Zinaida [17]

Answer:

W_{grav} < 0

Explanation:

When a projectile is fired upwards with some initial speed then the it reaches the top of the projectile and then falls back to the ground.

According to the question we need to find the work done by the gravity which is acting downwards for the projectile when it is at a position just about to hit the ground in course of falling down.

As we know that work is given as:

W=F.s\cos\theta

here:

F= force of gravity on the object (which is acting downwards)

s= displacement of the object

  • Here the work done by the gravity at an instant just before the projectile hits the earth will be negative as the displacement is in the direction opposite to the force of gravity.
7 0
4 years ago
Multiple-Concept Example 7 and Interactive LearningWare 26.1 provide some helpful background for this problem. The drawing shows
Vlada [557]

Answer:

n = 1.4266

Explanation:

Given that:

refractive index of crystalline slab n = 1.665

let refractive index of fluid is n.

angle of incidence θ₁ = 37.0°

Critical angle \theta _c = sin^{-1} (\frac{n}{n_{slab}} )

sin \theta _ c =\frac{n}{n_{slab}}

According to Snell's law of refraction:

n sin \theta _1 = n_{slab}  \ sin \  (90- \theta_c)

At point P ; 90 - \theta _2  \leq \theta _c

\theta _2 = 90 - \theta _c

Therefore:

n \ sin \theta_1 = n_{slab} \sqrt{(1-sin^2 \theta _c)}  \\ \\ n \ sin \theta_1 = n_{slab}   \sqrt{(1- \frac{n}{n_{slab}} )}

Then maximum value of refractive index  n of the fluid is:

n = \frac{n_{slab}}{\sqrt{1+ sin^2 \theta _1 } }

n = \frac{1.665}{\sqrt{1+ sin^2 \  37} }

n = 1.4266

3 0
3 years ago
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