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alekssr [168]
1 year ago
8

Light rays from stars bend toward smaller angles as they enter Earth's atmosphere. a. Explain why this happens using Snell's law

and the speed of light. b. Where are the actual stars in relation to their apparent position as viewed from the Earth's surface?

Physics
1 answer:
Grace [21]1 year ago
8 0

Answer:

Following are the answer to this question:

Explanation:

In option (a):

  • The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle.  
  • Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.

In option (b):

  • Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth.  
  • Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.

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The system shown above consists of two identical blocks that are suspended using four cords, each of a different length. Which o
Mariana [72]

Answer:

Option B and C are True

Note: The attachment below shows the force diagram

Explanation:

The weight of the two blocks acts downwards.

Let the weight of the two blocks be W. Solving for T₁ and T₂;

w = T₁/cos 60° -----(1)

w = T₂/cos 30° ----(2)

equating (1) and (2)

T₁/cos 60° = T₂/cos 30°

T₁ cos 30° = T₂ cos 60°

T₂/T₁ = cos 30°/cos 60°

T₂/T₁ =1.73

Therefore, option a is false since T₂ > T₁

Option B is true since T₁ cos 30° = T₂ cos 60°

Option C is true because the T₃ is due to the weight of the two blocks while T₄ is only due to one block.

Option D is wrong because T₁ + T₂ > T₃ by simple summation of the two forces, except by vector addition.

6 0
2 years ago
A water-skier with weight Fg = mg moves to the right with acceleration a. A horizontal tension force T is exerted on the skier b
Degger [83]

Answer:

The correct relationships are T-fg=ma and L-fg=0.

(A) and (C) is correct option.

Explanation:

Given that,

Weight Fg = mg

Acceleration = a

Tension = T

Drag force = Fa

Vertical force = L

We need to find the correct relationships

Using balance equation

In horizontally,

The acceleration is a

T-Fd=ma...(I)

In vertically,

No acceleration

w=L

mg-L=0

Put the value of mg

L-fg=0....(II)

Hence,  The correct relationships are T-fg=ma and L-fg=0.

(A) and (C) is correct option.

3 0
2 years ago
Dane is standing on the moon holding an 8 kilogram brick 2 metres above the ground. How much energy is in the brick's gravitatio
Nadya [2.5K]

The gravitational potential energy of the brick is 25.6 J

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its position in a gravitational field.

Near the surface of a planet, the gravitational potential energy is given by

PE=mgh

where

m is the mass of the object

g is the strength of the gravitational field

h is the height of the object relative to the ground

For the brick in this problem, we have:

m = 8 kg is its mass

g = 1.6 N/kg is the strenght of the gravitational field on the moon

h = 2 m is the height above the ground

Substituting, we find:

PE=(8)(1.6)(2)=25.6 J

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

3 0
1 year ago
Read 2 more answers
A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha
Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3  s

the formula for the time period of the pendulum is given by

T = 2\pi \sqrt{\frac{I}{mgd}}    .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

I = \frac{mL^{2}}{12}+ md^{2}

Substituting the values in equation (1)

3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}

9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )

12d² -26.84 d + 2.25 =  0

d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}

d=\frac{26.84\pm 24.75}{24}

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0
1 year ago
A traditional set of cycling rollers has two identical, parallel cylinders in the rear of the device that the rear tire of the b
mars1129 [50]

Answer:

ω2  =  216.47 rad/s

Explanation:

given data

radius r1 =  460 mm

radius r2 = 46 mm

ω =  32k rad/s

solution

we know here that power generated by roller that  is

power = T. ω    ..............1

power = F × r × ω

and this force of roller on cylinder is equal and opposite force apply by roller

so power transfer equal in every cylinder so

( F × r1 × ω1)  ÷ 2 = (  F × r2 × ω2 )  ÷  2    ................2

so

ω2  =  \frac{460\times 32}{34\times 2}

ω2  =  216.47

8 0
1 year ago
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