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Pachacha [2.7K]
3 years ago
13

Object A attracts object B with a gravitational force of 10 Newton’s from a given distance. If the distance between the two obje

cts is doubled, what is the new force of an attraction between them?
Physics
1 answer:
madam [21]3 years ago
7 0

If the distance between two objects is doubled, then the gravitational forces between them become (1/2)² = 1/4 of the original gravitational forces.

If the magnitude of those forces was 10 Newtons before the move then they're 2.5 Newtons after it.

I changed the wording of the discussion to plural, because the wording in the question is misleading.  It shouldn't say "Object A attracts Object B" and then let it go at that.  Gravitational forces ALWAYS occur in equal pairs.  If Object B feels a pull of 10 Newtons toward Object A, then Object A feels a pull of 10 Newtons toward Object B.  

The masses don't even matter.  Gravity is always there equally in both directions.  Your weight on Earth is equal to the Earth's weight on you !

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1. A 700 kg satellite is in orbit 2.4 x 106 meters from the center of the Earth. What is the force of
nevsk [136]

Answer:

4.8 \cdot 10^4 N

Explanation:

The force of gravity acting on the satellite is given by:

F=\frac{GMm}{r^2}

where

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

m is the mass of the satellite

r is the distance of the satellite from the Earth's centre

Here we have

m = 700 kg

r=2.4\cdot 10^6 m

Substituting into the equation, we find:

F=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(700)}{(2.4\cdot 10^6)^2}=4.8 \cdot 10^4 N

<em>Note that the distance mentioned in the problem (2.4 x 10^6 meters) is not realistic, since it is less than the radius of the Earth (6.37 x 10^6 meters).</em>

3 0
3 years ago
A 75 W lightbulb is being run on a 110 V outlet. Determine the resistance of the lightbulb. Provide a detailed description of th
Sauron [17]
Power is defined as

P = I*V

where I is the current and V is the voltage

Ohm's law gives us the relation betwen Voltage and current in a resistive component

V = I*R ,  Then

P = V² / R

We solve for R,

R = (110 V)²/ 75W =  161.33 ohms
8 0
3 years ago
What is the best structure for a egg dropping project you will be name brainiest
julsineya [31]

Answer:

bubble wrap in stuff animal

Explanation:

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4 0
2 years ago
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10
Svetlanka [38]

Answer:

D . A mass of 5 kilograms lifted 5 meters in 10 second

4 0
2 years ago
Calculate the equivalent resistance for both circuits. Series circuit: 2 Ω and 4 Ω Parallel circuit: 2 Ω and 4 Ω Which circuit h
goldenfox [79]
Equivalent resistance is also known as the overall resistance. 

For resistors in a series circuit, the total resistance is computed using the formula:

R_{T} = R_{1}+ R_{2}+ R_{3}... R_{n}

In other words, you just add up the resistance of each resistor in the series circuit. In your case you only have two resistors. You have 2Ω and 4Ω. So all you need to do is add that up. 

R_{T} = R_{1}+ R_{2}
R_{T} = 2 + 4=6

The total resistance of the series circuit is 6Ω

In a parallel circuit you get the total resistance using the formula:
\frac{1}{R_{T}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}...+\frac{1}{R_{n}}

First you get the sum of all fractions and at the end take the reciprocal of the resulting fraction and divide. So let us take your problem into consideration where you have two resistors that have a resistance of 2Ω and 4Ω.

\frac{1}{R_{T}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}
\frac{1}{R_{T}} = \frac{1}{2}+\frac{1}{4}
\frac{1}{R_{T}} = \frac{2}{4}+\frac{1}{4}
\frac{1}{R_{T}} = \frac{3}{4}

Get the reciprocal of the resulting fraction 3/4 and then divide. The reciprocal of 3/4 is 4/3.

4/3 = 1. 33Ω

So if you compare the equivalent resistance of the two circuits, the series circuit has a higher equivalent resistance. 
3 0
3 years ago
Read 2 more answers
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