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laiz [17]
3 years ago
6

Three identical stars of mass M form an equilateral triangle that rotates around the triangle’s center as the stars move in a co

mmon circle about that center. The triangle has an edge length of L. What is the speed of the stars?
Physics
1 answer:
zheka24 [161]3 years ago
5 0

Answer:

v=\sqrt{\dfrac{2GM}{L}}

Explanation:

M = Mass of planets

R = Radius of circle

v = Velocity

\theta = Angle

The circle is inside the triangle

cos\theta=\dfrac{\dfrac{L}{2}}{R}\\\Rightarrow R=\dfrac{L}{2cos\theta}

The centripetal acceleration

\dfrac{Mv^2}{R}=2\dfrac{GM^2}{L^2}cos\theta\\\Rightarrow \dfrac{Mv^2}{\dfrac{L}{2cos\theta}}=2\dfrac{GM^2}{L^2}cos\theta\\\Rightarrow \dfrac{Mv^22cos\theta}{L}=2\dfrac{GM^2}{L^2}cos\theta\\\Rightarrow v^2=\dfrac{2GM}{L}\\\Rightarrow v=\sqrt{\dfrac{2GM}{L}}

The speed of the stars is v=\sqrt{\dfrac{2GM}{L}}

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A runner accelerated to 4 m/s^2 for 20 seconds before winning the race.How far did he/she run?
creativ13 [48]

Answer:

s=800 m

Explanation:

Given that,

Acceleration of a runner, a = 4 m/s²

Time, t = 20 seconds

We need to find the distance covered by her. Initially, she was at rest. It means its initial velocity is equal to 0. So, using second equation of motion as follows :

s=ut+\dfrac{1}{2}at^2

Herre, u = 0

s=\dfrac{1}{2}at^2\\\\s=\dfrac{1}{2}\times 4\times (20)^2\\\\s=800\ m

So, she will cover a distance of 800 m.

8 0
3 years ago
A 25kg chair initially at rest on a horizontal floor requires a 165 N horizontal force to set it in motion. Once the chair is in
SCORPION-xisa [38]

Answer:

\mu_k=0.51  

Explanation:

Given that

Mass , m = 25 kg

We know that when body is in rest condition then static friction force act on the body and when body is in motion the kinetic friction force act on the body .That is why these two forces are given as follows

Static friction force ,fs= 165 N

Kinetic friction force ,fk = 127 N

If the body is moving with constant velocity ,it means that acceleration of that body is zero and all the forces are balanced.

Lets take coefficient of kinetic friction  = μk

The kinetic friction is given as follows

fk = μk  m g

Now by putting the values

127 = μk x 25 x 9.81

\mu_k=\dfrac{127}{25\times 9.81}

\mu_k=0.51

Therefore the value of coefficient of kinetic friction will be 0.51

4 0
3 years ago
The coach has not seen improvement as expected in his team and plans to change practice session. To help the players he changes
Allushta [10]

Answer:

progession

Explanation:

7 0
3 years ago
One hundred jumping beans are placed along the center line of a gymnasium floor at six-inch intervals. Twelve hours later, the d
stira [4]

Answer:

a) Diffusion  coefficient, D = 1.5 in/hr

b) Mean jump frequency, f = 0.0833 Hz

Explanation:

a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:

^{2} = 2Dt..........(1)

Where <r> = mean displacement

D = Diffusion coefficient

t = time = 12 hrs

sum of the squares of the distance divided by 100 is 36 in2.

<r>²= 36 in²

Substituting these values into equation (1) above

36 = 2 * D *12\\36 = 24 D\\D = 36/24\\D = 1.5 in/hr

b) Mean jumping distance, <r> = 0.1 inches

Applying equation (1) again

Where D = 1.5 in/hr

^{2} = 2Dt

0.1^{2}  = 2 * 1.5t\\0.01 = 3t\\t = 0.01/3\\t = 0.0033 hrs\\t = 0.0033 * 3600\\t = 12 seconds

The mean jump frequency, f = 1/t

f = 1/12

f = 0.0833 Hz

8 0
3 years ago
The intensity at a distance of 6.0 m from a source that is radiating equally in all directions is 6.0 × 10-10 w/m2 . what is the
satela [25.4K]
The intensity is defined as the ratio between the power emitted by the source and the area through which the power is calculated:
I= \frac{P}{A} (1)
where
P is the power
A is the area

In our problem, the intensity is I=6.0 \cdot 10^{-10} W/m^2. At a distance of r=6.0 m from the source, the area intercepted by the radiation (which propagates in all directions) is equal to the area of a sphere of radius r, so:
A=4 \pi r^2 = 4 \pi (6.0 m)^2 = 452.2 m^2

And so if we re-arrange (1) we find the power emitted by the source:
P=IA = (6.0 \cdot 10^{-10}W/m^2)(452.2 m^2)=2.7 \cdot 10^{-7} W
3 0
3 years ago
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