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shepuryov [24]
2 years ago
14

somewhere between the earth and the moon is a point where the gravitational attraction of the earth is canceled by the gravitati

onal pull of the moon. the mass of the moon is 1/81 that of the earth. how far from the center of the earth is this point?
Physics
1 answer:
mote1985 [20]2 years ago
8 0
<span>It's pretty easy problem once you set it up.

Earth------------P--------------Moon

"P" is where the gravitational forces from both bodies are acting equally on a mass m

Let's define a few distances.
Rep = distance from center of earth to P
Rpm = distance from P to center of moon
Rem = distance from center of earth to center of moon

You are correct to use that equation. If the gravitational forces are equal then

GMearth*m/Rep² = Gm*Mmoon/Rpm²

Mearth/Mmoon = Rep² / Rpm²

Since Rep is what you're looking for we can't touch that. We can however rewrite Rpm to be

Rpm = Rem - Rep

Mearth / Mmoon = Rep² / (Rem - Rep)²

Since Mmoon = 1/81 * Mearth
81 = Rep² / (Rem - Rep)²

Everything is done now. The most complicated part now is the algebra, so bear with me as we solve for Rep. I may skip some obvious or too-long-to-type steps.

81*(Rem - Rep)² = Rep²
81*Rep² - 162*Rem*Rep + 81*Rem² = Rep²
80*Rep² - 162*Rem*Rep + 81*Rem² = 0

We use the quadratic formula to solve for Rep:
Rep = (81/80)*Rem ± (9/80)*Rem
Rep = (9/8)*Rem and (9/10)*Rem

Obviously, point P cannot be 9/8 of the way to the moon because it'll be beyond the moon. Therefore, the logical answer would be 9/10 the way to the moon or B.

Edit: The great thing about this idealized 2-body problem, James, is that it is disguised as a problem where you need to know a lot of values but in reality, a lot of them cancel out once you do the math. Funny thing is, I never saw this problem in physics during Freshman year. I saw it orbital mechanics in my junior year in Aerospace Engineering. </span> sylent_reality · 8 years ago
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If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed?
sweet [91]

Answer:

Speed of the electron will be v=5.896\times 10^7m/sec

Explanation:

We have given that charge on electron e=1.6\times 10^{-19}C

Mass of electron m=9.11\times 10^{-31}kg

Potential difference = V=9.9KV=9.9\times 10^3volt

Now according to energy conservation eV=\frac{1}{2}mv^2

1.6\times 10^{-19}\times 9.9\times 10^3=\frac{1}{2}\times 9.11\times 10^{-31}v^2

v=5.896\times 10^7m/sec

4 0
1 year ago
Kenny and Candy decided to sit on a see-saw while visiting a local play park. Candy, of mass
pochemuha

Answer:

(i) 208 cm from the pivot

(ii) Move further from the pivot

Explanation:

(i) Sum of the moments about the pivot of the seesaw is zero.

∑τ = Iα

(50 kg) (10 N/kg) (2.5 m) + (60 kg) (10 N/kg) x = 0

1250 Nm + 600 N x = 0

x = -2.08 m

Kenny should sit 208 cm on the other side of the pivot.

(ii) To increase the torque, Kenny should move away from the pivot.

4 0
2 years ago
An individual white LED (light-emitting diode) has an efficiency of 20% and uses 1.0 W of electric power. a. How many LEDs must
Neporo4naja [7]

Answer:

8, 8 W

Explanation:

The useful power of 1 Light Emitting Diode is

0.2\times 1=0.2\ W

Total power required is 1.6 W

Number of Light Emitting Diodes would be

n=\dfrac{1.6}{0.2}\\\Rightarrow n=8

The number of Light Emitting Diodes is 8.

Power would be

P=8\times 1=8\ W

The power that is required to run the Light Emitting Diodes is 8 W

7 0
2 years ago
A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick sho
Gnoma [55]

Answer:34 cm

Explanation:

Given

mass of meter stick m=80 gm

stick is balanced when support is placed at 51 cm mark

Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark

balancing torque

80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)

80=5(50-x)

80=250-5x

5x=170

x=\frac{170}{5}

x=34 cm

4 0
2 years ago
You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 kg, was stopped at a red light w
Artemon [7]

Answer:

Explanation:

Force of friction at car B ( break was applied by car B ) =μ mg = .65 x  2100 X 9.8  = 13377 N .

work done by friction = 13377 x 7.30 = 97652.1 J

If v be the common velocity of both the cars after collision

kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²

= 1800 v²

so , applying work - energy theory ,

1800 v² = 97652.1

v² = 54.25

v = 7.365 m /s

This is the common velocity of both the cars .

To know the speed of car A , we shall apply law of conservation of momentum  .Let the speed of car A before collision be v₁ .

So , momentum before collision = momentum after collision of both the cars

1500 x v₁ = ( 1500 + 2100 ) x 7.365

v₁ = 17.676 m /s

= 63.63 mph .

( b )

yes Car A was crossing speed limit by a difference of

63.63 - 35 = 28.63 mph.

7 0
2 years ago
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