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Vladimir [108]
3 years ago
8

Compared to a stationary galaxy, light from a galaxy that is moving away from earth will appear _____.

Physics
2 answers:
Vlad1618 [11]3 years ago
3 0
"B" When an object moves away from us, the light is shifted to the red end of the spectrum, as its wavelengths get longer.
nikklg [1K]3 years ago
3 0

Answer:

B) Redder

Explanation:

Think of it like this; a car passes by you, as it goes, it's horn gets louder, and then starts to become quieter as it moves further away. Redshift works the same way. When an galaxy starts to move away from Earth, it's light <em>appears </em>to be redder. (If it was moving closer to Earth, it would most likely become bluer.)

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Red light is bent the least of all colors as it passes through a prism. What does this tell you about red light? It has a short
Alik [6]

Answer:

Longest wavelength, lowest intensity

Explanation:

7 0
3 years ago
A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel
Dafna1 [17]

Answer:

1.69\cdot 10^{10}J

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

E=-G\frac{mM}{2r}

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

M=5.98\cdot 10^{24}kg is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m

So the initial total energy is

E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J

When the satellite hits the ground, it is now on Earth's surface, so

r=R=6370 km=6.37\cdot 10^6 m

so its gravitational potential energy is

U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J

And since it hits the ground with speed

v=1.90 km/s = 1900 m/s

it also has kinetic energy:

K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J

So the total energy when the satellite hits the ground is

E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J

8 0
3 years ago
A shot-putter accelerates a 7.2 kg shot from rest to 17 m/s . what work did the shot-putter do on the ball?
garri49 [273]
<span>1.0x10^3 Joules The kinetic energy a body has is expressed as the equation E = 0.5 M V^2 where E = Energy M = Mass V = Velocity Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion E = 0.5 * 7.2 kg * (17 m/s)^2 E = 3.6 kg * 289 m^2/s^2 E = 1040.4 kg*m^2/s^2 E = 1040.4 J So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>
6 0
3 years ago
A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed pa
pav-90 [236]

Answer

given,

current (I) = 16 mA

circumference of the circular loop (S)= 1.90 m

Magnetic field (B)= 0.790 T

S = 2 π r

1.9 = 2 π r

r = 0.3024 m

a) magnetic moment of loop

    M= I A

    M=16 \times 10^{-3} \times \pi \times r^2

   M=16 \times 10^{-3} \times \pi \times 0.3024^2

   M=4.59 \times 10^{-3}\ A m^2

b)  torque exerted in the loop

 \tau = M\ B

 \tau = 4.59 \times 10^{-3}\times 0.79

 \tau = 3.63 \times 10^{-3} N.m

8 0
3 years ago
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