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2 years ago

Does the area of contact effect the frictional force . explain?

Physics

1 answer:

OlgaM077 [116]2 years ago

3 0

Answer:

The force due to friction is generally independent of the contact area between the two surfaces. This means that even if you have two heavy objects of the same mass, where one is half as long and twice as high as the other one, they still experience the same frictional force when you drag them over the ground.

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The energy band gap of GaAs is 1.4eV. calculate the optimum wavelength of light for photovoltaic generation in a GaAs solar cell

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Answer:

λ = 8.88 x 10⁻⁷ m = 888 nm

Explanation:

The energy band gap is given as:

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Converting this to Joules (J)

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Therefore,

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2 years ago

A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) Ho

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Answer:

a)N = 3.125 * 10¹¹

b) I(avg) = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

$N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1$

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

= (500)(0.5)(0.1 × 10⁻⁶)

= 2.5 × 10⁻⁵A

C) If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

$P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}$

= 1250W

d) Final peak=

P= Ik/e

= $= P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W$

P = 2.5 × 10⁷W

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Does the area of contact effect the frictional force . explain?​

Does the area of contact effect the frictional force . explain?