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kirill115 [55]
3 years ago
14

a 3000 kg and a 7000 kg Mass attract each other with a force of 0.0015 N. What distance separates the two objects (Radius) (Plea

se Show Steps)​
Physics
1 answer:
frutty [35]3 years ago
3 0

Answer:

<h3> 3.057m</h3>

Explanation:

According to law of gravitation;

F = GMm/d²

G is the universal gravitation

M and m are the masses

d is the distance between the masses

d² = GMm/F

d² = 6.67408 × 10-11 *3000*7000/0.0015

d² = 140.15568*10^-5/0.0015

d² = 1.4016*10^-3/0.0015

d² = 1.4016*10^-3/1.5*10^-3

d²  = 0.9344*10

d² = 9.344

d = √9.344

d = 3.057m

Hence the distance between the two objects is  3.057m

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How much work did the movers do (horizontally) pushing a 41.0-kg crate 10.6 m across a rough floor without acceleration, if the
VLD [36.1K]

Answer:

The required work done is 2555.448~J

Explanation:

Consider 'F' is the applied force on the crate and 'f' be the force created by friction. According to the figure if '\mu_{k}' be the coefficient of friction, then

f = \mu_{k} \times N = \mu_{k} \times Mg

where 'M', 'N' and 'g' are the mass of the crate, the normal force aced upon the block and the acceleration due to gravity respectively.

Since the application of force by the movers does not create any acceleration to the block, we can write

F = f = \mu_{k} \times M \times g = 0.6 \times 41~Kg~ \times 9.8~m~s^{-2} = 241.08~N

So the work done (W) in moving the crate by a distance s = 10.6 m is

W = F \times s = 241.08~N \times 10.6~m = 2555.448 J

5 0
3 years ago
Jonah observes the Sun through a special filtered telescope during a total solar eclipse. He sees a red ring and a faint white r
ivann1987 [24]

The answer would be C. chromosphere and corona.

8 0
3 years ago
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Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in
beks73 [17]

Answer:

a. the force between them quadruples

Explanation:

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

In this problem, the charges on both objects are doubled, so

q_1' = 2q_1\\q_2' = 2q_2

While the distance does not change, so the new force will be

F'= k \frac{(2q_1)(2q_2)}{r^2}=4 (k\frac{q_1 q_2}{r^2})=4 F

so, the force will quadruple.

4 0
3 years ago
Place the following types of electromagnetic radiation in order from LOWEST energy
Alex

Placing the electromagnetic radiation in order from the lowest energy to the highest energy :  ( 2 ) C,A,B

<h3>Electromagnetic spectrum </h3>

In the electromagnetic spectrum the electromagnetic radiation with the shorter wavelength possess a higher energy while the electromagnetic radiation with a longer wavelength possess the lower energy.

The electromagnetic radiation as listed in the question with the longest wavelength is the radio waves therefore it possess the lowest energy while the radiation with the shortest wavelength is the gamma rays therefore it possess the highest energy.

Hence we can conclude that Placing the electromagnetic radiation in order from the lowest energy to the highest energy :  ( 2 ) C,A,B

Learn more about electromagnetic spectrum : brainly.com/question/25847009

3 0
2 years ago
Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900
Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron m_{e}=9.11\times10^{-31}\ kg

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}

E=0.582\ Mev

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

E=0.350\ Mev

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0
3 years ago
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