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
alex41 [277]
2 years ago
11

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

are mA = 340 kg, mB = 567 kg, and mC = 139 kg. Take the positive direction to be to the right. Find the net gravitational force, including sign, acting on (a) particle A, (b) particle B, and (c) particle C.

Physics
1 answer:
miss Akunina [59]2 years ago
3 0

Formula of the gravitational force between two particles:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67 \cdot 10^{-11} Nm^2 kg^{-2} is the gravitational constant

m1 and m2 are the masses of the two particles

r is their distance


(a) particle A

The gravitational force exerted by particle B on particle A is

F_B=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=5.14 \cdot 10^{-5} N to the right

The gravitational force exerted by particle C on particle A is

F_C=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=5.6 \cdot 10^{-6} N to the right

So the net gravitational force on particle A is

F_A = F_B + F_C =5.14 \cdot 10^{-5} N+5.6 \cdot 10^{-6} N=5.7 \cdot 10^{-5} N to the right


(b) Particle B

The gravitational force exerted by particle A on particle B is

F_A=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=-5.14 \cdot 10^{-5} N to the left

The gravitational force exerted by particle C on particle B is

F_C=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=8.41 \cdot 10^{-5} N to the right

So the net gravitational force on particle B is

F_B = F_A + F_C =-5.14 \cdot 10^{-5} N+8.41 \cdot 10^{-5} N=3.27 \cdot 10^{-5} N to the right


(c) Particle C

The gravitational force exerted by particle A on particle C is

F_A=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=-5.6 \cdot 10^{-6} N to the left

The gravitational force exerted by particle B on particle C is

F_B=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=-8.41 \cdot 10^{-5} N to the left

So the net gravitational force on particle C is

F_C = F_B + F_A =-8.41 \cdot 10^{-5} N-5.6 \cdot 10^{-6} N=-8.97 \cdot 10^{-5} N to the left



You might be interested in
A white dwarf is:______.
bija089 [108]

Answer:

the exposed core of a dead star, supported by electron degeneracy pressure.

Explanation:

A white dwarf is a low luminosity exposed core of a dead star having mass comparable to the sun but volume comparable to the earth . So its density is very high . These stars have lost the capacity to generate energy through the process of fusion . Due to high gravitational energy , it goes on shrinking but ultimately balanced by electron degeneracy pressure. It is not a main sequence star as it has lost the power of fusion .

6 0
3 years ago
A motorcycle of mass 100 kilograms travels around a flat, circular track of radius 10 meters with a constant speed of 20 meters
JulijaS [17]

Answer:

100/10 = 10 , 10 × 10 = 100÷20 = 5

I'm pretty sure its wrong

8 0
3 years ago
Read 2 more answers
Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.3 m/s parallel to the gro
pashok25 [27]

Answer:

Explanation:

cSep 20, 2010

well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.

Anonymous

Sep 20, 2010

First you need to solve for time by using

d=(1/2)(a)(t^2)+(vi)t

1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s

t=.45 sec

Then you find the horizontal distance traveled by using

v=d/t

1.3m/s=d/.54sec

d=.585m

Then you need to find the time of player B by using

d=(1/2)(a)(t^2)+(vi)t

1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0

t=.61 sec

Finally to find player Bs initial horizontal velocity you use the horizontal equation

v=d/t

v=.585m/.61 sec

so v=.959m/s

5 0
3 years ago
Read 2 more answers
When you changed from low to high power, how did the change affect the working distance of the lens?
Basile [38]

The working distance gets shorter as the magnification gets bigger. In order to focus, the high-power objective lens must be significantly nearer to the specimen than the low-power lens. Magnification is negatively correlated with working distance.

Magnification change The magnification of a specimen is increased by switching from low power to high power. The magnification of an image is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens, or eyepiece.

The geometry of the optical system connects the magnifying power, or how much the thing being observed seems expanded, and the field of view, or the size of the object that can be seen.

To know more about  working distance

brainly.com/question/13551539

#SPJ4

4 0
1 year ago
Weight of 1 kg becomes 1/6 on moon. If radius of moon is 1.76×10^6 m, then the mass of moon will be​
igomit [66]

Answer:

7.65 x 10^22kg

sorry if im wrong!

5 0
2 years ago
Other questions:
  • Consider a 150-w incandescent lamp. the filament of the lamp is 5-cm long and has a diameter of 0.5 mm. the diameter of the glas
    10·1 answer
  • Which scale has 100 divisions from when the temperature when water freezes to the temperature when water boils
    13·1 answer
  • Use the following photoelectric graph to answer the following question:
    5·1 answer
  • A 0.35-kgkg cord is stretched between two supports, 7.4 mm apart. When one support is struck by a hammer, a transverse wave trav
    9·1 answer
  • When organic polymers joined together on early earth, they formed ______? Protons Protobionts Monomers Plankton
    14·2 answers
  • Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum
    11·1 answer
  • The frequency of a simple pendulum that makes 120 complete oscillations in 3 minutes is:
    12·1 answer
  • Please PLEASE HELP ASAP
    6·1 answer
  • Choose one of these three people and write a letter to them to express your concerns about the events of 01/06/21
    10·1 answer
  • Please help <br> Physics is so confusing
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!