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4 years ago

Work & Power Problems

Physics

1 answer:

Artyom0805 [142]4 years ago

6 0

Answer:

1. Unit for work is measured in Joules.

Formula for work is power * time taken

2 The unit for force is Newton

A work done in lifting a jar of water

3. The unit for measuring power is Joules.

The formula for measuring power is work done/time taken

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4 years ago

Which sequence occurs from fastest to slowest on the earth's surface?

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The answer should be A.

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3 years ago

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4 years ago

Please help on this one?

Rufina [12.5K]

Answer

D. The acceleration vector of the planet points towards the center of the circle.

Explanation

For planetary motion , the velocity vector is always in a direction that is tangential to the direction of motion, not towards the center. Even though the speed of the planet may stay constant, it's velocity changes because it is constantly changing it's direction.

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3 years ago

We have an initially uncharged hollow metallic sphere with radius of 5.0 cm. I place a small object with a charge of +20 µC at t

Advocard [28]

Answer:

d. 1.8E+7 N/C

Explanation:

In order to find the electric field outside the hollow metallic sphere, we can apply Gauss' Law, using a spherical gaussian surface with radius equal to 10 cm from the center of the sphere.

As the electric field must be normal to the surface at any point (no tangential fields can exist in electrostatic conditions) it must be radially pointed. By symmetry, at a same radius, the magnitude of the field must be the same.

As the dA vector is always normal to the surface and aiming outward, the dot product E*dA, can be taken out of the integral, as follows:

$E*A = \frac{Qenc}{\epsilon0}$

where Qenc, is the total charge enclosed by the gaussian surface. Just due to the conservation of charge, this charge must be equal to +20 μC.

Now, how can this charge be distributed on the outer surface of the shell?

If we apply Gauss´Law to a gaussian surface with a radius just inside the sphere (between the inner and outer surface), we will find that the flux is 0, due to the electric field is 0 inside a conductor.

We could write the same equation as above:

$E*A = \frac{Qenc}{\epsilon0} = 0$

If the left side of the equation is 0, the right one must be zero too:

⇒ Qenc = 0 ⇒ Qc + Qin = 0 ⇒ Qin = -20 μC.

As the metallic sphere must remain neutral, an equal and opposite charge must build up on the outer surface:

⇒ Qou = +20 μC

The other parameters in the equation are:

r = 0.1 m

ε₀ = 8.85*10⁻¹² C²/N*m²

Replacing by the values, we can solve for E as follows:

$E = \frac{1}{4*\pi*\epsilon0} *\frac{Qenc}{r^{2}} = \frac{1}{4*\pi*8.85e-12} *\frac{+20e-6C}{(0.1m)^{2}} = 1.8e7 N/C$

⇒ E = 1.8*10⁷ N/C

So, the statement d. is true.

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3 years ago