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netineya [11]
3 years ago
6

Far from any other masses, two masses, m1 and m2, are interacting gravitationally. The value for the mass of m1 suddenly doubles

. What happens to the value of the gravitational force that mass m2 exerts on mass m1?
Physics
1 answer:
NNADVOKAT [17]3 years ago
5 0
<span>It also doubles The gravitational force between two masses is expressed as: F = G*m1*m2/r^2 where F = Force between the two masses m1 = Mass of object 1 m2 = Mass of object 2 r = distance between centers of object 1 and object 2 G = Gravitational constant The exact values of G, m1, m2, and r don't matter since all except for m1 is held constant. And when m1 suddenly doubles, the force attracting the two object to each other also doubles.</span>
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What is the separation distance, in meters, between masses m1 = 15 x 107 kg and m2 = 62 x 107 kg when the gravitational force be
nalin [4]

Answer:

94.1 m

Explanation:

From Coulombs law,

F  = Gm1m2/r²................... Equation 1

where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.

Make r the subject of the equation,

r = √(Gm1m2/F)................. Equation 2

Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute into equation 2

r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)

r √(886.16×10)

r √(88.616×10²)

r = 9.41×10

r = 94.1 m.

Hence the distance of separation = 94.1 m

3 0
3 years ago
A ball hits a wall horizontally at 6m/s and rebounces at 4.4m/s the ball is in contact with wall for 0.04 sec. what is the accel
Svetlanka [38]

Answer:

Acceleration (a) = 40 m/s²

Explanation:

Given:

Initial velocity (u) = 6 m/s

Final velocity (v) = 4.4 m/s

Time taken (t) = 0.04sec

Find:

Acceleration (a) = ?

Computation:

We know that,

⇒ v = u + at

⇒ a = (v - u) / t

⇒ Acceleration (a) = (4.4 - 6) / 0.04

⇒ Acceleration (a) = (-1.6) / 0.04

Acceleration (a) = 40 m/s²

8 0
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2
sasho [114]

Answer:

The answer is 40 cm.

Explanation:

7 0
3 years ago
What do scientists use to determine the temperature of a star?
Nastasia [14]

Answer:

Measure the brightness of a star through two filters and compare the ratio of red to blue light. Compare to the spectra of computer models of stellar spectra of different temperature and develop an accurate color-temperature relation.

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3 years ago
A piece of clay sits 0.10 m from the center of a potter’s wheel. If the potter spins the wheel at an angular speed of 15.5 rad/s
finlep [7]

Answer:

a=24.025\ m/s^2    

Explanation:

Given that

Distance from the center ,r= 0.1 m

The angular speed ,ω = 15.5 rad/s

We know that centripetal acceleration is given as

a=ω² r

a=Acceleration

r=Radius

ω=angular speed

a=ω² r

Now by putting the values in the above equation we get

a=15.5^2\times 0.1\ m/s^2

a=24.025\ m/s^2

Therefore the acceleration of the clay will be a=24.025\ m/s^2.

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3 years ago
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