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ivanzaharov [21]
3 years ago
13

The electric potential a distance r from a small charge is proportional to what power of the distance from the charge?

Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
3 0

Answer: r^-1

Explanation:

V = (k × Q) / r, where V is electric potential, k is constant and r is distance from charge.

Since Q will not change, it can be considered a constant too.

So, we can write V ∝ 1/r

which is same as V ∝ r^-1

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Distinguish between a converging lens and a diverging lens.
andriy [413]

Answer:A converging lens is thickest in the middle and causes parallel light rays to converge through the focal point on the opposite side of the lens. A diverging lens is thinner in the middle and causes parallel light rays to diverge away from the focal point on the same side of the lens.

Explanation:

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3 years ago
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vodka [1.7K]
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3 years ago
A horizontal force of magnitude 30.2 N pushes a block of mass 3.50 kg across a floor where the coefficient of kinetic friction i
marshall27 [118]

Answer:

A) 89.39 J

B) 30.39J

C) 23.8 J

Explanation:

We are given;

F = 30.2N

m = 3.5 kg

μ_k = 0.646

d = 2.96m

ΔEth (Block) = 35.2J

A) Work done by the applied force on the block-floor system is given as;

W = F•d

Thus, W = 30.2 x 2.96 = 89.39 J

B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;

ΔEth = μ_k•mgd

Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J

Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.

Thus,

ΔEth = ΔEth (Block) + ΔEth (floor)

Thus,

ΔEth (floor) = ΔEth - ΔEth (Block)

ΔEth (floor) = 65.59J - 35.2J = 30.39J

C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;

W = K + ΔEth

Therefore;

K = W - ΔEth

K = 89.39 - 65.59 = 23.8J

3 0
3 years ago
3. Calculate the wavelength of wave that has a frequency of 4.75 x 1012Hz.
Klio2033 [76]
  • Frequency=v=4.75×10^12Hz
  • Wavelength=?

We know

\boxed{\sf \lambda=\dfrac{C}{V}}

\\ \sf\longmapsto \lambda=\dfrac{3\times 10^8ms^{-1}}{4.75\times 10^{12}s^{-1}}

\\ \sf\longmapsto \lambda=0.631\times 10^{-4}m

\\ \sf\longmapsto \lambda=6.31\times 10^{-5}m

4 0
3 years ago
What’s a real life example of relative motion?
vesna_86 [32]

Answer:

let's say you're on a bus going 50 km/hr, you are moving at a velocity of zero relative to the bus. however, relative to the ground you are moving at the same velocity as the bus.

Explanation:

physics

7 0
3 years ago
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