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Sliva [168]
3 years ago
13

Exoplanets (planets outside our solar system) are an active area of modern research. Suppose astronomers find such a planet that

has the same mass as Earth, but has a radius that is about 10% less. Roughly, what acceleration due to gravity would you expect if you were standing on the surface of this new planet?
Physics
1 answer:
aleksandrvk [35]3 years ago
3 0

Answer:

g' = 12.11 m/s^2

Explanation:

As we know that the acceleration due to gravity is given as

g = \frac{GM}{R^2}

now for earth we know that

\frac{GM}{R^2} = 9.81 m/s^2

now if on the surface of another planet we know that the mass is same as that the mass of Earth but radius is 10% less than the radius of Earth

so we have

r = 0.9 R

so we will have

g' = \frac{GM}{(0.9R)^2}

g' = \frac{GM}{0.81 R^2}

g' = 1.23 \times 9.81

g' = 12.11 m/s^2

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A flash of red light and a flash of blue light enter a glass cube perpendicular to its surface at the same time. after passing t
Wewaii [24]

A flash of red light and a flash of blue light enter a glass cube perpendicular to its surface at the same time. after passing through the block, the red light pulse exits first.

For any medium, other than vacuum, the index of refraction for red light is slightly lower (closer to 1 ) than that for blue light. This means that when light goes from vacuum (or air) into glass, the red light deviates from its original direction less than does the blue light. Also, as the light reemerges from the glass into vacuum (or air), the red light again deviates less than the blue light. If the two surfaces of the glass are parallel to each other, the red and blue rays will emerge traveling parallel to each other, but displaced laterally from one another.

what is refractive index?

The ratio between the speed of light in medium to speed in a vacuum is the refractive index. When light travels in a medium other than the vacuum, the atoms of that medium continually absorb and re-emit the particles of light, slowing down the speed light.

learn more about refractive index from here: brainly.com/question/28203787

#SPJ4

7 0
2 years ago
During a visit to the beach, you get in a small rubber raft and paddle out beyond the surf zone. Tiring, you stop and take a res
Monica [59]

Answer:

The main difference in these two movements is that the first is a pure swing movement and the followed form a wave travels from the beach

Explanation:

The movement in the two parts is very different, when the surf zone has passed it is in a deeper part of the water where the seabed does not rise much, therefore due to the movement of the waves there is an upward oscillatory movement and descending, in this movement there is no horizontal displacement.

When it is within the southern zone, there is a rapid rise of the sea floor, which generates a horizontal movement, having a traveling wave, therefore your movement is more complicated, you can have some oscillating movement on the axis and, but in addition to this you have a horizontal movement that reaches you towards the beach, forming a Traveling wave.

The main difference in these two movements is that the first is a pure swing movement and the followed form a wave travels from the beach

3 0
3 years ago
A 2.6 kg rock is dropped from a height of 10 m. With what speed will it strike the ground. Ignore air resistance. Solve using co
timurjin [86]

Answer:

mgh =  \frac{1}{2} m {v}^{2}  \\ v =  \sqrt{2gh}  \\  =  \sqrt{2 \times 9.8 \times 10}  \: m. {s}^{ - 1}

5 0
3 years ago
0.002 written in scientific notation
Oliga [24]

Answer:0,002 = 2 x 10⁻³

Explanation:

0,002 = 2 / 1000 = 2 / 10³ = 2 x 10⁻³

3 0
4 years ago
Huck Finn walks at a speed of 0.70 m/sm/s across his raft (that is, he walks perpendicular to the raft's motion relative to the
exis [7]

Answer:

Explanation:

Given

Velocity of Huck w.r.t to raft v_{H,raft}=0.7\ m/s

Perpendicular to the motion of raft

Velocity of Raft in the river v_{raft,river}=1.6\ m/s

As Huck is traveling Perpendicular to the raft so he possess two velocities i.e. vertical velocity and horizontal velocity of River when observed from bank

v_{Huck,river\ bank}=0.7\hat{j}+1.6\hat{i}

So magnitude of velocity is given by

|v|=\sqrt{0.7^2+1.6^2}

|v|=\sqrt{0.49+2.56}

|v|=\sqrt{3.05}

|v|=1.74\ m/s

For direction \tan =\frac{0.7}{1.6}=0.4375

\theta =23.63^{\circ} w.r.t river bank

                       

4 0
3 years ago
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