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erastovalidia [21]
3 years ago
13

A 65 kg student sits 3 m away from a 70 kg student. What is the magnitude of the gravitational force between the two students?

Physics
1 answer:
LiRa [457]3 years ago
3 0

Answer:

F=3.37\times 10^{-8}\ N

Explanation:

Given that,

Mass of student 1, m₁ = 65 kg

Mass of student 2, m₂ = 70 kg

The distance between the students, d = 3 m

We need to find the magnitude of the gravitational force between the two students. The formula for the magnitude of the gravitational force between two masses is given by :

F=G\dfrac{m_1m_2}{d^2}\\\\F=6.67\times 10^{-11}\times \dfrac{65\times 70}{(3)^2}\\F=3.37\times 10^{-8}\ N

So, the gravitational force between the two students is 3.37\times 10^{-8}\ N.

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Which of the following most logically completes the argument below?
kolezko [41]

Answer:

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Explanation:

The trade is one strong reflection for many weak reflections (and more dangerous near-Earth space travel).

None of the answer choices except the last one has anything to do with the effect of exploding a satellite. When you are arguing that exploding a satellite is ill conceived, you need to address specifically the effects of exploding the satellite.

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3 years ago
A basketball star covers 3.05 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled pr
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<span><span>anonymous </span> 4 years ago</span>Any time you are mixing distance and acceleration a good equation to use is <span>ΔY=<span>V<span>iy</span></span>t+1/2a<span>t2</span></span> I would split this into two segments - the rise and the fall. For the fall, Vi = 0 since the player is at the peak of his arc and delta-Y is from 1.95 to 0.890. For the upward part of the motion the initial velocity is unknown and the final velocity is zero, but motion is symetrical - it takes the same amount of time to go up as it does to go down. Physiscists often use the trick "I'm going to solve a different problem, that I know will give me the same answer as the one I was actually asked.) So for the first half you could also use Vi = 0 and a downward delta-Y to solve for the time. Add the two times together for the total. The alternative is to calculate the initial and final velocity so that you have more information to work with.
5 0
3 years ago
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What is the electric force acting between two charges of -0.0045 C and -0.0025 C that are 0.0060 m apart? Use Fe=kq1q2/r^2 and k
oksano4ka [1.4K]

Answer:

D. 2.8 × 10⁹ N

Explanation:

The force between two charges is directly proportional to the amount of charges at the two points and inversely proportional to the square of distance between the two points.

Fe= k Q₁Q₂/r²

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Q₂= -0.0025 C

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4 0
3 years ago
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Explanation:

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