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Taya2010 [7]
2 years ago
15

Consider two celestial objects with masses m1 and m2 with a separation distance between their centers r. If the first mass m1 we

re to double and the second mass m2 were to triple, what would happen to the magnitude of the force of attraction
Physics
1 answer:
Scilla [17]2 years ago
6 0

The new magnitude of the force of attraction will be 6 times the original force of attraction

<h3>How to determine the initial force </h3>
  • Mass 1 = m₁
  • Mass 2 = m₂
  • Gravitational constant = G
  • Distance apart = r
  • Initial force (F₁) = ?

F = Gm₁m₂ / r²

F₁ = Gm₁m₂ / r²

<h3>How to determine the new force </h3>
  • Mass 1 = 2m₁
  • Mass 2 = 3m₂
  • Gravitational constant = G
  • Distance apart (r) = r
  • New force (F₂) =?

F = Gm₁m₂ / r²

F₂ = G × 2m₁ × 3m₂ / r²

F₂ = 6Gm₁m₂ / r²

But

F₁ = Gm₁m₂ / r²

Therefore

F₂ = 6Gm₁m₂ / r²

F₂ = 6F₁

Thus, the new magnitude of the force of attraction will be 6 times the original force of attraction

Learn more about gravitational force:

brainly.com/question/21500344

#SPJ1

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Atomic physicists usually ignore the effect of gravity within an atom. To see why, we may calculate and compare the magnitude of
STatiana [176]

Answer:

2.27\cdot 10^{49}

Explanation:

The gravitational force between the proton and the electron is given by

F_G=G\frac{m_p m_e}{r^2}

where

G is the gravitational constant

m_p is the proton mass

m_e is the electron mass

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

F_G=(6.67259\cdot 10^{-11} m^3 kg s^{-2})\frac{(1.67262\cdot 10^{-27}kg) (9.10939\cdot 10^{-31}kg)}{(3 m)^2}=1.13\cdot 10^{-68}N

The electrical force between the proton and the electron is given by

F_E=k\frac{q_p q_e}{r^2}

where

k is the Coulomb constant

q_p = q_e = q is the elementary charge (charge of the proton and of the electron)

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

F_E=(8.98755\cdot 10^9 Nm^2 C^{-2})\frac{(1.602\cdot 10^{-19}C)^2}{(3 m)^2}=2.56\cdot 10^{-19}N

So, the ratio of the electrical force to the gravitational force is

\frac{F_E}{F_G}=\frac{2.56\cdot 10^{-19} N}{1.13\cdot 10^{-68}N}=2.27\cdot 10^{49}

So, we see that the electrical force is much larger than the gravitational force.

5 0
3 years ago
A particle (m = 4.3 × 10^-28 kg) starting from rest, experiences an acceleration of 2.4 × 10^7 m/s^2 for 5.0 s. What is its de B
Novay_Z [31]

Answer:

Wavelength, \lambda=1.28\times 10^{-14}\ m

Explanation:

Given that,

Mass of the particle, m=4.3\times 10^{-28}\ kg

Acceleration of the particle, a=2.4\times 10^7\ m/s^2

Time, t = 5 s

It starts from rest, u = 0

The De Broglie wavelength is given by :

\lambda=\dfrac{h}{mv}

v = a × t

\lambda=\dfrac{h}{mat}

\lambda=\dfrac{6.67\times 10^{-34}}{4.3\times 10^{-28}\times 2.4\times 10^7\times 5}

\lambda=1.28\times 10^{-14}\ m

Hence, this is the required solution.

4 0
3 years ago
A 1400 kg car is traveling east on the highway at 31 m/s and collides into the rear of a slower moving pickup truck of 2400 kg,
zloy xaker [14]

Answer: 31 m/s due east

Explanation: this question can be solved using the law of conservation of linear momentum.

This law states that in a closed or isolated system, during collision, the vector sum of momentum before collision equals the vector sum of momentum after collision.

Momentum = mass × velocity

From our question, our parameters before collision are given below as

Mass of car = mc = 1400kg

Speed of car =vc = 31 m/s (due east)

Mass of truck = mt = 2400kg

Velocity of truck = vt = 25 m/s ( due east )

After collision

Velocity of car = ?

Velocity of truck = 34 m/s ( due east )

Vector sum of momentum before collision is given as

1400 (31) + 2400 (25) = 43400 + 60000 = 103400 kgm/s

After collision the truck is seen to move faster (v = 34 m/s) which implies that the car also moves due east .

1400 (v) + 2400(25) .... A positive value is between both momenta because they are in the same direction.

After collision, we have that

1400v + 60000

Vector sum of momentum before collision = vector sum of momentum after collision

103400 = 1400v + 60000

103400 - 60000 = 1400v

43400 = 1400v

v = 43400/ 1400

v = 31 m/s due east

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Look at the graph pic and answer the question correctly!
SCORPION-xisa [38]

Answer:

B)

Explanation:

That the time period of which they stop.

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Explain why the speed of light is lower than<br> 3.0 x 108 m/s as it goes through different media.
jekas [21]

Answer:

Sample Response: In a vacuum, there are no atoms or particles that interfere with the path of light. However, in other media, the speed of light is lower than 3.0 × 108 m/s because the wave is continuously absorbed and re-emitted by each atom in its path. The differences in speed are due to the composition of the medium and the density of the particles in the medium.

Explanation:

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