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podryga [215]
3 years ago
12

The masses of the Earth and Moon are 5.98×1024kg and 7.35×1022kg respectively, and their centers are separated by 3.84×108m.

Physics
1 answer:
bija089 [108]3 years ago
4 0

Answer:

C) The Earth-Moon CM follows the orbit around the Sun. Earth and Moon rotate around their CM. The radius of rotation of Moon around the CM is much greater than radius of rotation of Earth around the CM.

Explanation:

At first we assume that both Earth and Moon can be treated as particles, the center of mass of the Earth-Moon system is obtained by using this formula:

r_{CM} = \frac{r_{E}\cdot m_{E}+r_{M}\cdot m_{M}}{m_{E}+m_{M}} (Eq. 1)

Where:

r_{E} - Location of the center of the Earth, measured in kilometers.

r_{M} - Location of the Moon, measured in kilometers.

r_{CM} - Location of the center of mass, measured in kilometers.

m_{E} - Mass of the Earth, measured in kilograms.

m_{M} - Mass of the Moon, measured in kilograms.

If we know that r_{E} = 0\,km, r_{M} = 3.84\times 10^{8}\,m, m_{E} = 5.98\times 10^{24}\,kg and m_{M} = 7.35\times 10^{22}\,kg, the location of the center of mass respect to the Earth is:

r_{CM} = \frac{(0\,km)\cdot (5.98\times 10^{22}\,kg)+(3.84\times 10^{8}\,m)\cdot (7.35\times 10^{22}\,kg)}{5.98\times 10^{24}\,kg+7.35\times 10^{22}\,kg}

r_{CM} = 4.662\times 10^{6}\,m

The Earth has a radius of 6.371\times 10^{6} meters, we notice that center of mass in located inside the Earth and the radius of rotation of the Earth around the center of mass is much greater than the radius of rotation of the Moon around the center of mass. That center of mass follows an orbit around the sun.

In consequence, correct answer is C.

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find the period of a simple pendulum of 1m length placed on earth and on moon g on moon =1.67m/s² g on earth=10m/s²
Ierofanga [76]

Answer:

T_{m } = 4.86 s

T_{e} = 1.98 s

Explanation:

<u><em>Given:</em></u>

Length = l = 1 m

Acceleration due to gravity of moon = g_{m} = 1.67 m/s²

Acceleration due to gravity of Earth = g_{e} = 10 m/s²

<u><em>Required:</em></u>

Time period = T = ?

<u><em>Formula:</em></u>

T = 2π \sqrt{\frac{l}{g} }

<u><em>Solution:</em></u>

<u>For moon</u>

<em>Putting the givens,</em>

T = 2(3.14) \sqrt{\frac{1}{1.67} }

T = 6.3 \sqrt{0.6}

T = 6.3 × 0.77

T = 4.86 sec

<u>For Earth,</u>

<em>Putting the givens</em>

T = 2π \sqrt{\frac{1}{10} }

T = 2(3.14) \sqrt{0.1}

T = 6.3 × 0.32

T = 1.98 sec

3 0
3 years ago
ow long must a simple pendulum be if it is to make exactly ten swings per second? (That is, one complete vibration takes exactly
Igoryamba
The period T of a pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the length of the pendulum while g=9.81 m/s^2 is the gravitational acceleration.

In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is T=0.200 s. Using this data, we can solve the previous formula to find L:
L=g ( \frac{T}{2\pi} )^2=(9.81 m/s^2)( \frac{0.2 s}{2 \pi} )^2=1 \cdot 10^{-3} m=1 mm
4 0
3 years ago
If the angular frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum acceleration
Nataly_w [17]

Answer:

When we double the angular velocity the maximum acceleration (a_{max}) will changes by a factor of 4.

Explanation:

Given the angular frequency (\omega) of the simple harmonic oscillator is doubled.

We need to find the change in the maximum acceleration of the oscillator.

a_{max}=A\omega^2

Now, according to the problem, the angular frequency (\omega) got doubled.

Let us plug \omega=2\times \omega. Then the maximum acceleration will be a_{max'}

a_{max}=A\omega^2

a_{max'}=A(2\times \omega)^2\\a_{max'}=A\times 4\omega\\a_{max'}=4A\omega

a_{max'}=4a_{max}

We can see, when we double the angular velocity the maximum acceleration will changes by a factor of 4.

6 0
3 years ago
What is the average acceleration of a tennis ball that has an initial velocity of 6.0 m/s [E] and a final velocity of 7.3 m/s [W
Marizza181 [45]

Given :

The average acceleration of a tennis ball that has an initial velocity of 6.0 m/s.

and a final velocity of 7.3 m/s.

It is in contact with a tennis racket for 0.094 s

To Find :

The average acceleration of the tennis ball.

Solution :

We know, average acceleration is given by :

a_{avg}=\dfrac{Final \ velocity-Initial\ velocity}{Time\ Taken}\\\\a_{avg}=\dfrac{7.3-6.0}{0.094}\ m/s^2\\\\a_{avg}=13.83\ m/s^2

Therefore, average velocity is given by 13.83 m/s².

Hence, this is the required solution.

7 0
2 years ago
Canadian geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at
Wewaii [24]

Answer:

θ=19.877⁰

Explanation:

Given data

Velocity Va=34.0 km/h

Velocity Va=100 km/h

To find

Angle θ

Solution  

We want the bird to fly with velocity Vb=100 km/h with an angle θ relative to the ground so that the bird fly due south relative to the ground.From figure which is attached we got

Sinθ=(Va/Vb)

Sinθ=(34.0/100)

θ=Sin⁻¹(34.0/100)

θ=19.877⁰

5 0
3 years ago
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