1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Eddi Din [679]
2 years ago
12

(b) The distance of mass from mass A if there is no gravitational force acted on C

Physics
1 answer:
shepuryov [24]2 years ago
7 0

Answer:

(a) The force, acting on object 'C' is approximately 2.66972 × 10⁻¹⁰ Newtons

(b) The distance of 'C' from 'A', in the direction particle 'B' if there is no  meters gravitational force acting on 'C' is appromimately 0.829 meters or 1.877 meters

Explanation:

The given parameters are;

The mass of particle, A, m₁ = 2 kg

The mass of particle, B, m₂ = 0.3 kg

The mass of particle, C, m₃ = 0.05 kg

The distance between particle 'A' and particle 'B', r₁ = 0.15 m

The distance between particle 'B' and particle 'C', r₂ = 0.05 m

(a) The gravitational force, 'F', is given as follows;

F =G \times \dfrac{m_{1} \times m_{2}}{r^{2}}

Where;

F = The force between the two masses

G = The gravitation constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m₁ = The mass of object 1

m₂ = The mass of object 2

If 'C' is placed at 0.05 m from 'B', we have;

F₂₃ =  6.67430 × 10⁻¹¹ × 0.05 × 0.3/(0.05²) ≈ 4.00458 × 10⁻¹⁰

The gravitational force between force between particle 'B' and particle 'C', F₂₃ = 4.00458 × 10⁻¹⁰ N (towards the right)

F₁₃ =  6.67430 × 10⁻¹¹ × 0.05 × 2/(0.1²) ≈ × 10⁻¹⁰

The gravitational force between force between particle 'A' and particle 'B', F₁₃ = 6.6743 × 10⁻¹⁰ N (towards the left)

The force, 'F', acting on object 'C' = F₁₃ - F₂₃

F = (6.6743 - 4.00458) × 10⁻¹⁰ = 2.66972 × 10⁻¹⁰ N

The force, acting on object 'C' ≈ 2.66972 × 10⁻¹⁰ N

(b), When there is no gravitational force acting on 'C', let the distance of 'C' from 'A' = x

We have;

F₂₃ = F₁₂

F_{23} =G \times \dfrac{m_{1} \times m_{2}}{r_1^{2}} = F_{13} =G \times \dfrac{m_{1} \times m_{3}}{r_2^{2}}

By plugging in the values and removing like terms, we get;

\dfrac{0.3 \times 0.05}{(1.15 - x)^{2}}  = \dfrac{2 \times 0.05}{x^2}

(1.15 - x)² × 2 × 0.05 = 0.3 × 0.05 × x²

0.1·x² - 0.23·x + 1.3225 = 0.015·x²

0.1·x² - 0.23·x + 1.3225 - 0.015·x² = 0

0.085·x² - 0.23·x + 0.13225= 0

x = (0.23± √((-0.23)² - 4 × 0.085 × ( 0.13225)))/(2 × 0.085))

x ≈ 0.829, or x ≈ 1.877

Therefore, the distance of 'C' from 'A', if there is no gravitational force acting on 'C', x ≈ 0.829 m, or x = 1.877 m, in the direction of 'B'

You might be interested in
What is the momentum of a 45-kg quarterback moving eastward at 15<br> m/s?
puteri [66]

Answer:

Given

mass (m) =45kg

velocity (v) =15m/s

momentum (p) =?

Form

p=mv

=45x 15

p=675kg.m/s

the momentum =675kg.m/s

4 0
3 years ago
Researchers conducted an experiment to identify the effects of three different hand sanitizers on the growth of bacteria. The re
pickupchik [31]
Picture ? I need a visual reference
7 0
3 years ago
Read 2 more answers
A closed path that allows electrical energy to flow is called a(n)
Bumek [7]

Answer:

Circuit

Explanation:

4 0
2 years ago
Read 2 more answers
A 25.0 kg pickle is accelerated from rest through a distance of 6.0m in 4.0s across a level floor . If the friction force betwee
SIZIF [17.4K]
Add the KE increase and the work done against friction.

The final velocity is twice the average, or 3.0 m/s
The final KE is (1/2)*25*3^2 = 112.5 J

The friction work done is 6*3.8 = 22.8 J 
 hope this is correct
8 0
3 years ago
A child of mass m is standing at the edge of a carousel. Both the carousel and the child are initially stationary. The carousel
suter [353]

Answer:

the angular velocity of the carousel after the child has started running =

\frac{2F}{mR} \delta t

Explanation:

Given that

the mass of the child = m

The radius of the disc = R

moment of inertia I = \frac{1}{2} mR^2

change in time = \delta \ t

By using the torque around the inertia ; we have:

T = I×∝

where

R×F = I × ∝

R×F = \frac{1}{2} mR^2∝

F = \frac{1}{2} mR∝

∝ = \frac{2F}{mR}           ( expression for angular  angular acceleration)

The first equation of motion of rotating wheel can be expressed as :

\omega = \omega_0  + \alpha  \delta t

where ;

∝ = \frac{2F}{mR}    

Then;

\omega = 0+ \frac{2F}{mR} \delta t

\omega =  \frac{2F}{mR} \delta t

 

∴ the angular velocity of the carousel after the child has started running =

\frac{2F}{mR} \delta t

7 0
3 years ago
Other questions:
  • Find the voltage across and power absorbed by the 25 ohm resistor.
    12·1 answer
  • An 8.00 kg point mass and a 15.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be
    15·1 answer
  • Why is the gravitational potential energy of an object 1 meter above the moon’s surface less than its potential energy 1 meter a
    10·1 answer
  • Please Help! 30 points! ASAP PLZ :)
    8·2 answers
  • Which of the following solutions should be chosen to help restore an ecosystem naturally? A. reintroducing an animal to the ecos
    6·1 answer
  • Where is the electric field closest to uniform? inside a circle described by a point charge at its center between parallel plate
    15·1 answer
  • Consider two diffraction gratings with the same slit separation. The only difference between the two gratings is that one gratin
    12·1 answer
  • A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 41.0 lb/in2. If the pump is a cylinder of lengt
    5·1 answer
  • Russell drags his suitcase 15.0 M from the door of his house to the car at a constant speed with a horizontal force of 95.0 N. H
    12·1 answer
  • Hello people ~
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!