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3 years ago

When you apply the torque equation ∑τ = 0 to an object in equilibrium, the axis about which torques are calculated:

Physics

1 answer:

kupik [55]3 years ago

3 0

Answer:

option D.

Explanation:

The correct answer is option D.

When an object is in equilibrium torque calculated at any point will be equal to zero.

An object is said to be in equilibrium net moment acting on the body should be equal to zero.

If the net moment on the object is not equal to zero then the object will rotate it will not be stable.

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In a 2-dimensional coordinate system, which of these is an example of an object's position?

Serjik [45]

' B ' is the only choice on the list that tells you a definite place, where you could go to pick up the object or meet somebody.

But in order to use it, you also have to know where the origin of coordinates is ... the point (0, 0) .

__________

No, I'm wrong about that. You have to know where SOME other point is, but that doesn't have to be (0, 0).

4 0

3 years ago

What is kinetic energy

pogonyaev

Answer:

Kinetic energy is the energy of motion

6 0

3 years ago

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A spherical bowling ball with mass m = 4.1 kg and radius R = 0.117 m is thrown down the lane with an initial speed of v = 8.9 m/

weqwewe [10]

Answer:Given mass = 4.1kg

Radius = 0.0117m

Velocity V = 8.4m/s

Coefficient of friction = 0.25

Explanation: Below is an attached solution to the problem stated above.

1. The angular acceleration is equal to 524rad/s^2

2. The linear acceleration is equal to 2.45m/s^2

3.the time it takes the ball to begin rolling = 0.98s

4. The distance the ball slides before it begins to roll = 7.05m

3 0

4 years ago

A boy lifted a 50 newton rock 1 meter. How much work was done?

dem82 [27]

Answer:

The answer is 50 Nm

Explanation:

<h3><u>Given</u>;</h3>

Applied Force = 50 Newton
Total Displacement = 1 meter

Work done = ?

Here,

W = F • d

W = 50 • 1

W = 50 Nm

Thus, Work done is 50 Nm

<u>-TheUnknownScientist 72</u>

5 0

3 years ago

Read 2 more answers

The radius of the earth's very nearly circular orbit around the sun is 1.5 1011 m. find the magnitude of the earth's velocity, a

kotykmax [81]

The radius of Earth's circular trajectory is
$R=1.5*10^{11} m$
The time for the Earth to travel around the Sun is
$t=365(days)*24(hours)*3600(seconds) =3.1536*10^7 (s)$
Thus the velocity is velocity
$v=2\pi R/t =2\pi *1.5*10^{11}/(3.1536*10^{7})=2.99*10^4 (m/s)$
From this we can deduce the centripetal acceleration
$a=v^2/R =(2.99*10^4)^2/(1.5*10^{11}) =5.96*10^{-3} (m/s^2)$
Angular velocity is
$\omega =v/R =(2.99*10^4)/(1.5*10^{11})=2*10^{-7} (rad/s)$

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4 years ago