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fredd [130]
4 years ago
13

2. 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:
algol [13]4 years ago
3 0

Explanation:

a)Snell's law states that when light travels from a rarer to a denser substance, like air to water or from a less dense layer of the atmosphere to a denser layer, it bends towards the normal{an imaginary line that is perpendicular to the surface of both media}. However,the opposite occurs when light moves from a more dense to a less dense medium. The angle between the normal and the refracted light ray is known as the angle of refraction.

In case of earth as light from the stars enters the earth atmosphere it bends towards smaller angle because the earth density increases as the light travel towards the earth troposphere from the exosphere as per the Snell's law described above.

b)Light rays that travel straight down do not bend, while rays that enter the Earth's atmosphere at a shallower angle get refracted and bend towards the normal, roughly following the direction of the Earth's curvature.

This means that celestial objects in the zenith position directly above you appear in the correct position, while objects closer to the horizon appear to be higher up in the sky than they actually are.

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An angry rhino with a mass of 2600 kg charges directly toward you with a speed of 3.70 m/s. Before you start running, as a distr
worty [1.4K]

Answer:

vf₂ = 15.79 m/s

Explanation:

Theory of collisions  

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:  

p=m*v  

where  

p:Linear momentum  

m: mass  

v:velocity  

There are 3 cases of collisions : elastic, inelastic and plastic.  

For the three cases the total linear momentum quantity is conserved:  

P₀ = Pf  Formula (1)  

P₀ :Initial linear momentum quantity  

Pf : Final linear momentum quantity  

Data

m₁ = 2600 kg : mass of the rhino

m₂= 0.18 kg : mass of the ball

v₀₁ = 3.70 m/s  : initial velocity of the rhino

v₀₂= - 8.39 m/s, initial velocity of the ball

Problem development

We appy the formula (1):

P₀ = Pf  

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂  

(2600)*(3.7) + (0.18)*(- 8.39) = (2600)*vf₁ +(0.18)*vf₂

9620 -1.5102 = (2600)*vf₁ +(0.18)*vf₂

9618.4898 = (2600)*vf₁ +(0.18)*vf₂ Equation (1)

Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.

e = \frac{v_{f2}-v_{f1} }{v_{o1}-v_{o2}}

1*( v₀₁- v₀₂ )  = (vf₂ -vf₁)

( 3.7 -( -8.39 )  = (vf₂ -vf₁)

12.09 =  (vf₂ -vf₁)

vf₂ = vf₁ + 12.09 Equation (2)

We replace Equation (2) in the Equation (1)

9618.4898 = (2600)*vf₁ +(0.18)*vf₂

9618.4898 = (2600)*vf₁ +(0.18)* (vf₁ + 12.09)

9618.4898 = (2600)*vf₁ +(0.18)*vf₁ + (0.18)(12.09)

9618.4898 = (2600)*vf₁ +(0.18)*vf₁ + 2.1762

9618.4898 -2.1762 = (2600.18)*vf₁

9616.3136 = (2600.18)*vf₁  

vf₁ = (9616.3136) / (2600.18)

vf₁ = 3.698 m/s : Final velocity of the rhino

We replace vf₁ = 3.698 m/s in the Equation (2)

vf₂ = vf₁ + 12.09

vf₂ = 3.698 + 12.09

vf₂ = 15.79 m/s : Final velocity of the ball

6 0
3 years ago
Label these as balanced or unbalanced:
kvasek [131]

Answer:

  1. A book lying on a table  - Balanced force
  2. An airplane cruising in level flight  - Balanced
  3. A rock falling from a cliff  - Unbalanced force
  4. A bridge collapsing in an earthquake  -  Unbalanced force
  5. A man sitting on a park bench  - Balanced force
  6. A space shuttle taking off  - Unbalanced force
  7. A car maintaining a constant speed on a straight road  - Balanced force
  8. An airplane landing - Unbalanced force

Explanation:

Usually, one or more forces act on a body at an instant of time. When these forces acting on a body and bring the body in the equilibrium position, the force is said to be balanced. The unbalanced force changes the equilibrium state of the body.

As in the case of an airplane cruising in a level flight, the weight of the plane will be equal to the lift force and the thrust is equal to the drag. So the plane is experiencing a balanced force.

4 0
3 years ago
Read 2 more answers
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Art [367]
Uh I think it’s point guard
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3 years ago
Which formula can be used to find the tangential speed of an orbiting object?
Luden [163]
The correct formula for calculating the tangential speed of an orbiting object is V(t)=wr.
V(t)= Tangential Speed
w= Angular Velocity
r= Radius of the Path

Hope this helps.
6 0
3 years ago
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A system consists of N particles that can occupy two energy levels: a nondegenerate ground state and a three-fold degenerate exc
cupoosta [38]

Answer:

Ng = 0.893 N,  Ne = 0.107N

Explanation:

Number of particles in Ground state = Ng

Number of particles in Excited state = Ne

Ne/Ng = e^{(-ΔE)/kt}

Since excited state is 3 fold degenerate

Ne/Ng =3 x e^{(-ΔE)/kt}

ΔE = Energy difference between ground and excited states = 0.25eV

T = 960 K

Constant k = 8.617 x 10^-5 eV/K

Ne/Ng = 3 x e^{-0.25/(8.617x10^-5) x 960}

           = 3 x e^(-3.188645)

           = 3 x 0.0412 = 0.1237 ≅ 0.12

Ne = 0.12 Ng

but Ne + Ng = N, where is N is total number of particles, substituting Ne into equation we get,

Ng(1 + 0.12) = N

Ng = N/1.12 = 0.893N

and Ne = 0.12 x 0.893 N = 0.107 N

3 0
4 years ago
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