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2 years ago

An uncharged, nonconducting, hollow sphere of

Physics

1 answer:

Aleonysh [2.5K]2 years ago

3 0

Answer:

Explanation:

The whole surface of hollow sphere = 4π r²

= 4 x 3.14 x (10 x 10⁻²)²

= 12.56 x 10⁻² m²

Area of the hole ( both side ) = 2 x π r²

= 2 x 3.14 x (10⁻³)²

= 6.28 x 10⁻⁶ m²

flux coming out of given charge at the centre as per Gauss's theorem

= q / ε₀ where q is charge at the centre and ε₀ is permittivity of the medium .

= 10 x 10⁻⁶ / 8.85 x 10⁻¹²

= 1.13 x 10⁶

This flux will pass through the surface of sphere so flux passing through per unit area

= 1.13 x 10⁶ / 12.56 x 10⁻²

= 8.99 x 10⁶ weber per m²

flux through area of hole

= 8.99 x 10⁶ x 6.28 x 10⁻⁶

= 56.45 weber .

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The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming a

Vera_Pavlovna [14]

Answer:

The radius of the gold nucleus is 7.1x10⁻¹⁵m

Explanation:

The nearest distance is:

$r=\frac{kze^{2} }{mpV^{2} }$ (eq. 1)

Where

z = atomic number of gold = 79

e = electron charge = 1.6x10⁻¹⁹C

k = electrostatic constant = 9x10⁹Nm²C²

energy of the particle = 32 MeV = 5.12x10⁻¹²J

At the potential energy is zero, all the energy will be kinetic energy:

$E_{k} =\frac{1}{2} m V^{2}$

Where

m = 4 mp = mass of proton

$5.12x10^{-12} =\frac{1}{2} *4*m_{p}* V^{2} \\m_{p}* V^{2} = 2.56x10^{-12}$

Replacing in equation 1

$r=\frac{9x10^{9}*79*(1.6x10^{-19})^{2} }{2.56x10^{-12} } =7.1x10^{-15} m$

8 0

2 years ago

4.Give at least three examples from everyday life where an inclined plane is used to reduce the effort force needed to accomplis

olga nikolaevna [1]

1. A wheelchair ramp. Instead of using lifting force on the wheelchair, You use push or pull force on it.

2. A slide. Instead of throwing down an item, It uses gravitational potential energy make an object "move" down the slide.

3.A screw. It's reducing the force by twisting the screw out of something instead of pulling it out. (Sorry about my bad grammar).

5 0

3 years ago

A boy pushes a cart with a constant velocity of 0.5m/s by applying a force of 60 N. What is the total frictional force acting on

Nutka1998 [239]

Answer:

60N

Explanation:

in this case the minimum amount of force required must be equal to the friction Force. i.e Newton's first law of motion.

therefore the maximum amount of frictional force is equal to the applied force which is 60N.

because of the net force acting on the object is zero the object is in constant motion . i.e equal and opposite force must be applied so that the object is in constant velocity therefore the total frictional force must be 60N

8 0

3 years ago

As a way of determining the inductance of a coil used in a research project, a student first connects the coil to a 5.62 V batte

Reptile [31]

Answer:

Its inductance L = 166 mH

Explanation:

Since a current, I = 0.698 A is obtained when a voltage , V = 5.62 V is applied, the resistance of the coil is gotten from V = IR

R = V/I = 5.62/0.698 = 8.052 Ω

Since we have a current of I' = 0.36 A (rms) when a voltage of V' = 35.1 V (rms) is applied, the impedance Z of the coil is gotten from

V₀' = I₀'Z where V₀ = maximum voltage = √2V' and I₀ = maximum current = √2I'

Z = V'/I' = √2 × 35.1 V/√2 × 0.36 V = 97.5 Ω

WE now find the reactance X of the coil from

Z² = X² + R²

X = √(Z² - R²)

= √(97.5² - 8.05²)

= √(9506.25 - 64.8025)

= √9441.4475

= 97.17 Ω

Now, the reactance X = 2πfL where f = frequency of generator = 93.1 Hz and L = inductance of coil.

L = X/2πf

= 97.17/2π(93.1 Hz)

= 97.17 Ω/584.965 rad/s

= 0.166 H

= 166 mH

Its inductance L = 166 mH

5 0

3 years ago

How much power is generated if 375 J of work is done in 15 s?

erma4kov [3.2K]

Answer: 25 watt

Explanation:

Given,

Work= 375 j

Time= 15

So, power = W÷t

= 375 ÷ 15

= 25 watt

7 0

2 years ago