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mr_godi [17]
2 years ago
12

What is the tension in the string if the turntable is held fixed, with the mass hanging from the pulley (i. e. , the mass is sta

tionary)?
Physics
1 answer:
Paladinen [302]2 years ago
7 0

Weight of the mass is the tension in the string if the turntable is held fixed, with the mass hanging from the pulley

Weight is a force that acts at all times on all objects near Earth. The Earth pulls on all objects with a force of gravity downward toward the center of the Earth. Weight (symbolized w ) is a quantity representing the force exerted on a particle or object by an acceleration field, particularly the gravitational field of the Earth at the surface.

The tension force is defined as the force that is transmitted through a rope, string or wire when pulled by forces acting from opposite sides. The tension force is directed over the length of the wire and pulls energy equally on the bodies at the ends.

The direction of tension is the pull which is given the name tension. Thus, the tension will point away from the mass in the direction of the string/rope. In case of the hanging mass, the string pulls it upwards, so the string/rope exerts an upper force on the mass and the tension will be in the upper side.

Tension will act opposite the the weight of the mass attached in order to maintain equilibrium .

To learn more about weight here

brainly.com/question/2293244

#SPJ4

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Answer:

|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}

Now, substitute this into 'dF':

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)

Now we can integrate dF over the rod.

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