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Arisa [49]
3 years ago
12

Use the following information about Earth and its moon to determine the distance between them. Earth's mass = 6.0 x 1024 kg Moon

's tangential speed = 1,025 m/s Earth's radius = 6.4 x 10 m Moon's mass = 7.3 X 1022 kg 4.6 x 106 m O 3.8 x 108 m O 3.9 x 1011 m​
Physics
2 answers:
MrRissso [65]3 years ago
6 0

Answer:

the planet Uranus has mass 8.7 x 1025 kg. ... radius 5.1 x 107 m around Uranus. G= 6.7 x ... The gravitational potential of the satellite is its centripetal force. ... Moon. Use this idea to calculate the mass of the Earth M from the values of vand rgiven ... 9mm . Ve GM. Ma var. 1023** 3-8* 10". M= ... 5.9x1024... kg [2]. 6-7x 10-11.

9966 [12]3 years ago
5 0

Answer:

3.8 × 108 m

Explanation:

Use formula: v^2 = G x m central / r

G = 6.77 e-11

Also, it was right on Edge.

Good luck!

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3 0
3 years ago
in the circuit diagram even in suppose the resistor R1 R2 and R3 have the value of 5 ohm 10 ohm 30 ohm respectively which have b
Anuta_ua [19.1K]
<h3><u>Given</u> :</h3>

Three identical resistors of resistances 5Ω, 10Ω and 30Ω are connected with a battery of 12V.

<h3><u>To Find</u> :</h3>

We have to find current through the each resistor and equivalent resistance of circuit.

<h3><u>SoluTion</u> :</h3>

➝ Equivalent resistance of series connection is given by

  • <u>R = R1 + R2 + R3</u>

➝ We know that, Equal current flow through each resistor in series connection.

➝ As per ohm's law, Current flow through a conductor is directly proportional to the applied potential difference.

  • <u>V = IR</u>

◈ <u>Equivalent resistance</u> :

⇒ Req = R1 + R2 + R3

⇒ Req = 5 + 10 + 30

⇒ <u>Req = 45Ω</u>

◈ <u>Current flow in circuit</u> :

⇒ V = IReq

⇒ 12 = I × 45

⇒ <u>I = 0.27A</u>

፨ Therefore, 0.27A current will flow through each resistor.

5 0
3 years ago
All of the following are locations for kinesthetic system receptors except __________. A. muscles B. semicircular canals C. join
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Answer:

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5 0
3 years ago
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A tuning fork labeled 392 Hz has the tip of each of its two prongsvibrating with an amplitude of 0.600mma) What is the maximum s
DanielleElmas [232]

a) 1.48 m/s

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v_{max}=\omega A

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v_{max} is the maximum speed

\omega is the angular frequency

A is the amplitude

For the tuning fork in the problem, we have

\omega=2\pi f=2 \pi(392 Hz)=2462 rad/s, where f is the frequency

A=0.600 mm=6\cdot 10^{-4} m is the amplitude

Therefore, the maximum speed is

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b)  3.0\cdot 10^{-5} J

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where

m=0.0270 g=2.7\cdot 10^{-5} kg is the mass of the fly

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Substituting into the equation, we find

K=\frac{1}{2}(2.7\cdot 10^{-5}kg)(1.48 m/s)^2=3.0\cdot 10^{-5} J

7 0
3 years ago
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