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fredd [130]
3 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]3 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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Mamont248 [21]

Answer:

Work and Kinetic Energy

A B

3. A 0.180 kg balls falls 2.5 m. How much work does the force of gravity do on the ball? 4.41 J

4. A forklift raises a box 1.2 m doing 7.0 kJ of work on it. What is the mass of the box? 595.24 kg

5. How much work does the force of gravity do when a 25 N object falls a distance of 3.5 m? 87.5 J

Explanation:

5 0
2 years ago
A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image
Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}

=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}

Re-arrange to get i

\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

m=\frac{h'}{h} =-\frac{i}{o}

substitute the values to get the height of image h'

\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm

5 0
4 years ago
In which medium does sound travel the fastest?
Vikki [24]

1.  D - sound travels the fastest through solids

2. 50 mm/s - v=fa

3. B - only process that involves changing waves

4 0
3 years ago
Help please asap
Rasek [7]
B because the mantle is farther down in and a egg has more of the hard stuff inside
8 0
3 years ago
One end of a spring with a force constant of k = 10.0 N/m is attached to the end of a long horizontal frictionless track and the
koban [17]

Answer:

-2.478

0.379

11.14

24.78

Explanation:

Angular frequency of spring in harmonic motion is given by?

ω = √(k/m)

ω = √(10/2.2)

ω = √4.54

ω = 2.13 s^-1

If at t=0 the mass is in negative amplitude (x = -A = -2.48 m) then we describe the position with negative cosine

x(t) = -A * cos(ωt)

x(t) = -2.48 * cos(2.13 * 1)

x(t) = -2.48 * 0.9993

x(t) = -2.478

Velocity and acceleration are 1st and 2nd derivative of position

b)

v(t) = Aω * sin(ωt)

v(t) = 2.48 * 2.13 * sin(2.13 * 1)

v(t) = 5.282 * sin2.13

v(t) = 5.282 * 0.03717

v(t) = 0.379 m/s

c)

a(t) = Aω^2 * cos(ωt)

a(t) = 2.48 * 2.12² * cos(2.13 * 1)

a(t) = 2.48 * 4.494 * cos2.13

a(t) = 11.15 * 0.9993

a(t) = 11.14 m/s²

d)

F = -k * x(t)

F = -10 * -2.478

F = 24.78 N

3 0
3 years ago
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