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9966 [12]
4 years ago
6

Where is the center of mass of the Earth-Moon system? The radius of the Earth is 6378 km and the radius of the Moon is 1737 km.

Select one of the answers below: Choose the correct description of the location of the center of mass of the Earth-Moon system. Choose the correct description of the location of the center of mass of the Earth-Moon system. The center of mass is exactly in the center between the Earth and the Moon. The center of mass is nearer to the Moon than the Earth, but outside the radius of the Moon. The center of mass is nearer to the Earth than the Moon, but outside the radius of the Earth.
Physics
1 answer:
lions [1.4K]4 years ago
3 0

Answer:

r_{cm} = 4.67 10⁶ m

We see that none of the answers is correct, the closest is the third, but the center of mass is within the radius of the Earth

Explanation:

The definition of the center of mass is

      r_{cm} = 1 / M ∑r_{i}  m_{i}

Where M is the total mass, r_{i} and m_{i}mi are the position and the mass gives each of the bodies

We must locate a reference system for the calculation, we will locate it with the origin in the center of the Earth, the data we have are

Mass of the Earth Me = 5.98 1024 kg

Moon mass m = 7.36 1022 kg

Earth to Moon Distance r = 3.84 108 m

Let's apply this to our case

      r_{cm} = 1 / (Me + m) (Me 0 + m R)

      r_{cm} = 1 / (598 +7.36) 10²² (0 + 7.36 10²² 3.84 10⁸)

     r_{cm} = 4.67 10⁶ m

We can see that this distance is less than the radius of the Earth

We see that none of the answers is correct, the closest is the third, but the center of mass is within the radius of the Earth

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Why are we seeing extremely old light from Canopus instead of light in real-time?
Irina-Kira [14]

Answer:

Canopus is more than 300 light years away from earth. This means it takes the light we see more than 300 years to reach us.

8 0
3 years ago
A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an an
frozen [14]

Answer:

a) <em>2.278 x 10^-5 volts</em>

b) <em>1.139 x 10^-6 Ampere</em>

c) <em>2.59 x 10^-11 W</em>

Explanation:

The radius of the wire r = 2 mm = 0.002 m

the number of turns N = 200 turns

direction of the magnetic field ∅ = 25°

magnetic field strength B = 0.02 T

varying time = 2 sec

The cross sectional area of the wire = \pi r^{2}

==> A = 3.142 x 0.002^{2} = 1.257 x 10^-5 m^2

Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°

==> Φ = 2.278 x 10^-7 Wb

The induced EMF is given as

E = NdΦ/dt

where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7

E = 200 x 1.139 x 10^-7 = <em>2.278 x 10^-5 volts</em>

<em></em>

<em></em>

b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as

I = E/R

where R is the resistor

I = (2.278 x 10^-5)/20 = <em>1.139 x 10^-6 Ampere</em>

<em></em>

<em></em>

<em> </em>c) power delivered to the resistor is given as

P = IE

P = (1.139 x 10^-6) x (2.278 x 10^-5) = <em>2.59 x 10^-11 W</em>

4 0
3 years ago
Free points to those who rate this 5 stars and answer "Done". I don't think it lets more than 2 people answer so I guess the fir
chubhunter [2.5K]

Answer:

how do i rate it 5 stars?

Explanation:

8 0
3 years ago
Read 2 more answers
A proton is moving in a circular orbit of radius 20 cm under a uniform magnetic field 0.3 t perpendicular to the velocity of the
Vladimir [108]

Answer:

v = 5.75 x 10⁶ m/s

Explanation:

The radius (r) of the circular orbit taken by a charged particle is related to its speed perpendicular to a magnetic field of strength B, and is given by

r = \frac{mv}{qB}       --------------(i)

Where,

q = charge of the particle

m = mass of the particle

Making v subject of the formula in equation (i) above gives

v = \frac{qBr}{m}  -------------------(ii)

Given;

r = 20cm = 0.2m

B = 0.3T

v = unknown

q = charge of proton = 1.6 x 10⁻¹⁹ C

m = mass of the proton = 1.67 x 10⁻²⁷kg

Substitute the values of m, q, B and r into equation (ii) above to get;

v = \frac{1.6 * 10^{-19} * 0.3 * 0.2} {1.67*10^{-27} }

Solving for v gives:

v = 5.75 x 10⁶ m/s

Therefore, the velocity of the proton is 5.75 x 10⁶ m/s

4 0
3 years ago
Which of the following is an example of adhesion?
nalin [4]
What are the answers?
3 0
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