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3 years ago

A crane lifts up two boxes. Which free body diagram shows the forces acting on block A?

Physics

2 answers:

Temka [501]3 years ago

7 0

Answer:

The 3rd graph

Explanation:

A free body diagram is a diagram which shows all the forces acting on an object.

The problem asks us to find the free body diagram of block A, so we must find all the forces acting on block A.

We have 3 forces acting on block A in total:

- The force of gravity (its weight), which pushes the block downward (in the diagram, it is the force represented with $F_{gA}$

- The tension in the rope 1, which pulls block A upwards: this force is represented with $F_{T1}$

- The tension in the rope 2, due to the weight of block 2, which pulls block A downwards: this force is represented with $F_{T2}$

Based on the direction of these 3 forces, the correct diagram is the 3rd one.

tatuchka [14]3 years ago

4 0

Answer:

t he third graph

Explanation:

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Answer:

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What is E(r)E(r)E(r), the radial component of the electric field between the rod and cylindrical shell as a function of the dist

andriy [413]

Answer:

E(r) = λ/2πrε0

Explanation:

If we consider an infinitely long line of charge with the charge per unit length being λ, we can take advantage of the cylindrical symmetry of this situation.

By symmetry, i mean that the electric fields all point radially away from the line of charge and thus there is no component parallel to the line of charge.

Niw, let's use a cylinder (with an arbitrary radius (r) and length (l)) centred on the line of charge as our Gaussian surface.

Doing that will mean that the electric field would be perpendicular to the curved surface of the cylinder. Therefore, the angle between the electric field and area vector is equal to zero and cos θ = cos 0 = 1

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Now, we know that according to Gauss Law,

Electric Flux, Φ = E•dA

Thus,

Total Φ = Φ_curved + Φ_top + Φ_bottom

Thus,

Φ = ∫E•dA cos 0 + ∫E•dA cos 90° + ∫E•dA cos 90°

We now have ;

Φ = ∫E . dA × 1

Since we are dealing with the radial component, the curved surface would be equidistant from the line of charge and the electric field in the surface will be the same magnitude throughout.

Thus,

Φ = ∫E•dA = E∫dA = E•2πrl

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q_net = λl

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Φ = E•2πrl = q_net/εo = λl/εo

E•2πrl = λl/ε0

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