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Diano4ka-milaya [45]
3 years ago
8

The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the

moon and that the mass of the earth is 5.98 x 10^24 kg. Find the period for the moon's motion around the earth. Express the answer in days and compare it to the length of a month.
Physics
1 answer:
Inga [223]3 years ago
6 0

Answer:

27.5 days

0.92 month

Explanation:

r = radius of the orbit of moon around the earth = 3.85\times10^{8} m

M = Mass of earth = 5.98\times10^{24} m

T = Time period of moon's motion

According to Kepler's third law, Time period is related to radius of orbit as

T^{2} = \frac{4\pi ^{2} r^{3}  }{GM}

inserting the values, we get

T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3}  }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec

we know that

1 day = 24 hours = 24 x 3600 sec = 86400 s

T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days

1 month = 30 days

T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month

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10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????
KATRIN_1 [288]

Answer: Tension = 47.8N, Δx = 11.5×10^{-6} m.

              Tension = 95.6N, Δx = 15.4×10^{-5} m

Explanation: A speed of wave on a string under a tension force can be calculated as:

|v| = \sqrt{\frac{F_{T}}{\mu} }

F_{T} is tension force (N)

μ is linear density (kg/m)

Determining velocity:

|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }

|v| = \sqrt{0.00874 }

|v| = 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

\Delta x = |v|.t

\Delta x = 9.35.10^{-2}*1.23.10^{-3}

Δx = 11.5×10^{-6}

With tension of 47.8N, a pulse will travel Δx = 11.5×10^{-6}  m.

Doubling Tension:

|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }

|v| = \sqrt{2.0.00874 }

|v| = \sqrt{0.01568}

|v| = 0.1252 m/s

Displacement for same time:

\Delta x = |v|.t

\Delta x = 12.52.10^{-2}*1.23.10^{-3}

\Delta x = 15.4×10^{-5}

With doubled tension, it travels \Delta x = 15.4×10^{-5} m

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What is potential energy?
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bit.^{}

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The rolling resistance for steel on steel is quite low; the coefficient of rolling friction is typically μr=0.002. Suppose a 180
Irina18 [472]

Answer:

t = 0.354 hours

Explanation:

given,

coefficient of rolling friction μr=0.002

mass of locomotive = 180,000 Kg

rolling speed = 25 m/s

The force of friction = μ mg

                                 = (.002) x (180000) x (9.8)

                                 = 3528 N

F = m  a

now,

m a =  3528 N

180000 x a = 3528

a = 0.0196 m/s²

Then apply

v = u + at  

0 = 25 - 0.0196 x t

t = 1275.51 sec

t = 1275.61/3600 hours

t = 0.354 hours

time taken by the locomotive to stop = t = 0.354 hours

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3 years ago
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