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gulaghasi [49]
3 years ago
14

In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point

ing electric field of 1.92 ✕ 105 N/C is applied. Find the charge on the drop, in terms of e.
Physics
1 answer:
Misha Larkins [42]3 years ago
7 0

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ  X volume of the oil drop

volume of the oil drop (spherical) =  (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

=  (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

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A plane flying against the wind covers the 900-kilometer distance between two aerodromes in 2 hours. The same plane flying with
Mashcka [7]

Answer:

The speed of the wind is 25 km/hr.

Explanation:

Let us call v_p the speed of the plane and v_w the speed of the wind. When the plane is flying against the wind, it covers the distance of 900-km in 2 hours (120 minutes); therefore;

(1). v_p - v_w = \dfrac{900km}{120min}

And when the plane is flying with the wind, it covers the same distance in 1 hour 48 minutes (108 minutes)

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From equation (1) we solve for v_p and get:

v_p = \dfrac{900km}{120min}+v_w,

and by putting this into equation (2) we get:

\dfrac{900km}{120min}+v_w+v_w= \dfrac{900km}{108min}

2v_w= \dfrac{900km}{108min}-\dfrac{900km}{120min}

2v_w = 8.3km/min - 7.5km/min

2v_w = 0.83km/min

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\boxed{v_w= 25km/hr }

4 0
3 years ago
A plane wall of thickness 0.1 mm and thermal conductivity 25 W/m K having uniform volumetric heat generation of 0.3 MW/m3 is ins
juin [17]

Answer:

The maximum temperature is 90.06° C

Explanation:

Given that

t= 0.1 mm

Heat generation

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Heat transfer coefficient

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Here one side(left side) of the wall is insulated so the all heat will goes in to right side .

The maximum temperature will at the left side.

Lets take maximum temperature is T

Total heat flux ,q

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q=0.3\times 1000000\times 0.1 \times 10^{-3}\ W/m^2

q=30\ W/m^2

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R=0.002 K/W

We know that

q=ΔT/R

30=(T-90)/0.002

T=90.06° C

The maximum temperature is 90.06° C

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Answer:

the answer is c

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