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Vesna [10]
3 years ago
5

What is the torque about the center of the sun due to the gravitational force of attraction of the sun on the planet?

Physics
1 answer:
Anon25 [30]3 years ago
3 0
Torque, moment, or moment of force is the tendency of a force to rotate an object around an axis,fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist to an object. Mathematically, torque is defined as the cross product of the vector by which the force's application point is offset relative to the fixed suspension point (distance vector) and the force vector, which tends to produce rotational motion. So torque about the center of the sun due to the gravitational force of attraction of the sun on the planet = (Gxm1xm2 / r^2). r sin(theta) = Gxm1xm2 /r^2). r sin0° = 0
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Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point
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a) 2.33 m/s

b) 5.21 m/s

c) 882 m³

Explanation:

Using the concept of continuity equation

for flow through pipes

A_{1}\times V_{1} = A_{2}\times V_{2}

Where,

A = Area of cross-section

V = Velocity of fluid at the particular cross-section

given:

A_{1} = 0.070 m^{2}

V_{1} = 3.50 m/s

a) A_{2} = 0.105 m^{2}

substituting the values in the continuity equation, we get

0.070\times 3.50 = 0.105\times V_{2}

or

V_{2} = \frac{0.070\times 3.5}{0.150}m/s

or

V_{2} = 2.33m/s

b) A_{2} = 0.047 m^{2}

substituting the values in the continuity equation, we get

0.070\times 3.50 = 0.047\times V_{2}

or

V_{2} = \frac{0.070\times 3.5}{0.047}m/s

or

V_{2} = 5.21m/s

c) we have,

DischargeQ = Area (A)\times Velocity(V)

thus from the given value, we get

Q = 0.070m^{2}\times 3.5m/s\

Q = 0.245 m^{3}/s

Also,

DischargeQ = \frac{volume}{time}

given time = 1 hour = 1 ×3600 seconds

substituting the value of discharge and time in the above equation, we get

0.245m^{3}/s = \frac{volume}{3600s}

or

0.245m^{3}/s\times 3600 = Volume

volume of flow = 882 m^{3}

8 0
3 years ago
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