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Genrish500 [490]
3 years ago
5

A. The distance to a star is approximately 8.58 ✕ 10^18 m. If this star were to burn out today, in how many years would we see i

t disappear?
b. How long does it take sunlight to reach Earth?
c. How long does it take for a microwave radar signal to travel from Earth to the Moon and back?
Physics
1 answer:
Tamiku [17]3 years ago
7 0

Answer:

906.27931 years

8.34 minutes

2.56266 seconds

Explanation:

v = Velocity of light = 3\times 10^8\ m/s

s = Distance

Time is given by

t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{8.58\times 10^{18}}{3\times 10^8\times 365.25\times 24\times 3600}\\\Rightarrow t=906.27931\ years

We would see the star disappear after 906.27931 years after it actually disappeared.

Distance between Earth and Sun = 150.18 billion m

t=\dfrac{150.18\times 10^9}{3\times 10^8\times 60}\\\Rightarrow t=8.34\ minutes

It takes 8.34 minutes for sunlight to reach Earth.

Distance between Earth and Moon = 384.4 million m

t=\dfrac{2\times 384.4\times 10^6}{3\times 10^8}\\\Rightarrow t=2.56266\ s

It takes 2.56266 seconds for a microwave radar signal to travel from Earth to the Moon and back.

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Answer:

\boxed{\sf Tension \ in \ the \ string \ (T) = 3 \ kN}

Given:

Mass (m) = 3.0 kg

Uniform speed (v) = 20 m/s

Length of string (r) = 40 cm = 0.4 m

To Find:

Tension in the string (T)

Explanation:

Tension (T) is the string will be equal to centripetal force (\sf F_c).

\boxed{ \bold{ T = F_c  =  \frac{m {v}^{2} }{r} }}

Substituting value of m, v & r in the equation:

\sf \implies T =  \frac{3 \times  {20}^{2} }{0.4}  \\  \\  \sf \implies T = \frac{3 \times 400}{0.4}  \\  \\  \sf \implies T =3 \times 1000 \\  \\  \sf \implies T =3000 \: N \\  \\ \sf \implies T =3 \: kN

\therefore

Tension in the string (T) = 3 kN

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The heat extracted from the cold reservoir Q cold (i.e., inside a refrigerator) divided by the work W required to remove the heat is known as the coefficient of performance, or COP, of a refrigerator (i.e., the work done by the compressor). The required inside temperature and the outside temperature have a significant impact on the COP.

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Refrigerator's coefficient of performance = H/W

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Therefore, the refrigerator's coefficient of performance is 6.

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