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9966 [12]
3 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]3 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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A parallel-plate capacitor has a plate area of 0.2m^2 and a plate separation of 0.1mm. To obtain an electric field of 2.0 × 10^6
Oduvanchick [21]

Answer:

3.536*10^-6 C

Explanation:

The magnitude of the charge is expresses as Q = CV

C is the capacitance of the capacitor

V is the voltage across the capacitor

Get the capacitance

C = ε0A/d

ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m

A is the area = 0.2m²

d is the plate separation = 0.1mm = 0.0001m

Substitute

C = 8.84 x 10-12 * 0.2/0.0001

C = 1.768 x 10-8 F

Get the potential difference V

Using the formula for Electric field intensity

E = V/d

2.0 × 10^6  = V/0.0001

V = 2.0 × 10^6  * 0.0001

V = 2.0 × 10^2V

Get the charge on each plate.

Q = CV

Q =  1.768 x 10-8 * 2.0 × 10^2

Q = 3.536*10^-6 C

Hence the magnitude of the charge on each plate should be 3.536*10^-6 C

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A worker drives a 0.562 kg spike into a rail tie with a 2.26 kg sledgehammer. The hammer hits the spike with a speed of 64.4 m/s
zubka84 [21]

Answer:

Explanation:

Given that,

Mass of sledge hammer;

Mh =2.26 kg

Hammer speed;

Vh = 64.4 m/s

The expression fot the kinetic energy of the hammer is,

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K.E(hammer) = ½ × 2.26 × 64.4²

K.E ( hammer) = 4686.52 J

If one forth of the kinetic energy is converted into internal energy, then

ΔU = ¼ × K.E(hammer)

∆U = ¼ × 4686.52

∆U = 1171.63 J

Thus, the increase in total internal energy will be 1171.63 J.

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3 years ago
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