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Aneli [31]
3 years ago
12

A binary star system consists of two stars of masses m1 and m2. The stars, which gravitationally attract each other, revolve aro

und the center of mass of the system. The star with mass m1 has a centripetal acceleration of magnitude a1. Note that you do not need to understand universal gravitation to solve this problem.Find a2, the magnitude of the centripetal acceleration of the star with mass m2.
Physics
1 answer:
zubka84 [21]3 years ago
8 0

Answer:

 a₂ = m₁ / m₂ a₁

Explanation:

For this exercise we note that the attraction between the two stars is an action and reaction force, therefore it has the same magnitude, but it is applied to each of the bodies

Let's apply Newton's second law on the star 1

     F₁ = m₁ a₁

Newton's second law in star 2

       F₂ = m₂ a₂

       | F₁ | = | F₂ |

      m₁  a₁ = m₂  a₂

      a₂ = m₁ / m₂ a₁

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1. its must be B and 2. must be C
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3 years ago
Two automobiles traveling at right angles to each other collide and stick together. Car A has a mass of 1200 kg and had a speed
sergij07 [2.7K]

Answer:

v_{B0}=15.73 m/s

Explanation:

We can use the conservation of momentum. The initial momentum is equal to the final momentum:

x-coordinate

p_{0x}=p_{fx}

m_{A}v_{A0}=(m_{A}+m_{B})v_{cx}  

m_{A}v_{A0}=(m_{A}+m_{B})v_{c}cos(40) (1)

y-coordinate

p_{0y}=p_{fy}

m_{B}v_{B0}=(m_{A}+m_{B})v_{cy}  

m_{B}v_{B0}=(m_{A}+m_{B})v_{c}sin(40) (2)

We can divide equations (2) and (1):

\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=\frac{sin(40)}{cos(40)}

\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=tan(40)

v_{B0}=\frac{m_{A}v_{A0}}{m_{B}}*tan(40)

v_{B0}=\frac{1200*25}{1600}*tan(40)

v_{B0}=15.73 m/s

I hope it helps you!

           

4 0
3 years ago
Read 2 more answers
If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object paral
leonid [27]

Answer:

\theta = tan^{-1}\mu

Explanation:

As we know that if the object is placed on the inclined plane then the force of friction on the object is counterbalanced by the component of the weight of the object along the inclined plane.

So we can say

F_f = mgsin\theta

now if we increase the inclination of the plane then the component of the weight weight along the inclined plane will increase and hence the friction force will also increase.

As we know that the limiting value or the maximum value of friction force at the static condition is given by

F_f = \mu N

N = mg cos\theta

so we have

F_f = \mu (mg cos\theta)

so we will have

mg sin\theta = \mu (mg cos\theta)

so now we have

tan\theta = \mu

so maximum possible angle of the inclined plane is

\theta = tan^{-1}\mu

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3 years ago
The field-line representation of the e-field in a certain region in space is shown below. The dashed lines represent equipotenti
lorasvet [3.4K]

You need to go on google an dlook this up

3 0
3 years ago
3. How much work is done when you pull a 6 N wagon for 5 meters?
8090 [49]

Answer:

<h2>30 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question

force = 6 N

distance = 5 m

We have

workdone = 6 × 5 = 30

We have the final answer as

<h3>30 J</h3>

Hope this helps you

3 0
2 years ago
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