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vladimir2022 [97]
3 years ago
8

4. Is the force of the earth’s gravity on the sun stronger, weaker, or the same as the force of the sun’s gravity on the earth?

Explain why the sun’s attraction affects the motion of the earth more than the earth’s attraction affects the sun’s motion. Employ Newton’s second and third laws in your answer.
Physics
1 answer:
Nady [450]3 years ago
5 0

Answer:

The first statement is false, the Sun has a stronger gravitational pull.

:

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That is force.  Whenever you see the words push or pull always think of Force
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The units of work,energy and power are...............units​
Andreas93 [3]

unit of work is joules

power - watt

energy - joules

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E asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 6.0 earth years. part a wha
anzhelika [568]

To solve the problem, use Kepler's 3rd law : 

T² = 4π²r³ / GM 

Solved for r : 

r = [GMT² / 4π²]⅓ 

but first covert 6.00 years to seconds : 

6.00years = 6.00years(365days/year)(24.0hours/day)(6... 
= 1.89 x 10^8s 

The radius of the orbit then is : 

r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓ 
= 6.23 x 10^11m

6 0
3 years ago
You weigh 730N. What would you weigh if the earth were six times as massive as it is and its radius were four times its present
Ber [7]

The new weight is 273.8 N

Explanation:

The weight of an object on Earth is given by

W=mg

where

m is the mass of the object

g is the acceleration of gravity

The acceleration of gravity at the Earth's surface can be rewritten as

g=\frac{GM}{R^2}

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

Substituting into the previous equation,

W=\frac{GMm}{R^2}

We said that in normal conditions, the weight of the person is 730 N:

W=\frac{GMm}{R^2}=730 N

Later, we are said that:

  • The mass of the Earth increases by a  factor of 6, M'=6M
  • The radius of the Earth increases by a factor of 4, R'=4R

Substituting into the equation, we find the new weight of the person in these conditions:

W'=\frac{G(6M)m}{(4R)^2}=\frac{6}{16}(\frac{GMm}{R^2})=\frac{3}{8}W

So, the new weight is 3/8 of the original weight, therefore:

W'=\frac{3}{8}(730)=273.8 N

Learn more about gravity:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

3 0
3 years ago
A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to
Nostrana [21]

Answer:

a) r=4.24cm d=1 cm

b) Q=5x10^{-10} C

Explanation:

The capacitance depends only of the geometry of the capacitor so to design in this case knowing the Voltage and the electric field

V=1.00x10^{2}v\\E=1.00x10^{4} \frac{N}{C}

V=E*d\\d=\frac{V}{E}\\d=\frac{1.0x10^{2}}{1.0x10^{4}}\\d=0.01m

The distance must be the separation the r distance can be find also using

C=\frac{Q}{V_{ab}}

But now don't know the charge these plates can hold yet so

a).

d=0.01m

C=E_{o}*\frac{A}{d}\\A=\frac{C*d}{E_{o}}

A=\frac{5pF*0.01m}{8.85x10^{-12}\frac{F}{m}}\\A=5.69x10^{-3}m^{2}

A=\pi *r^{2}\\r=\sqrt{\frac{A}{r}}\\r=\sqrt{\frac{5.64x10^{-3}m^{2} }{\pi } }  \\r=42.55x^{-3}m

b).

C=\frac{Q}{V_{ab}}

Q=C*V\\Q=5x10^{-12} F*1x10^{2}\\Q=5x10^{-10}C

5 0
3 years ago
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