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SpyIntel [72]
3 years ago
13

Calculate the electric field strength required to just support an oil drop of mass

Physics
1 answer:
Ray Of Light [21]3 years ago
3 0

Answer:

E= 3.06 \times 10^{-17} N/C

Explanation:

As we know that the oil drop experiment tells us the balancing the gravitational force by electrostatic force.

So ,

Electrostatic force = Gravitational force

qE=mg

Here , q is charge , E is electric field , m is mass and g is gravity.

so,

E=\frac{mg}{q}

Insert values from question,

E=\frac{10^{-2}kg \times 9.8 m/s^{2}}{3.2 \times 10^{-19}C}

or

E= 3.06 \times 10^{-17} N/C

This is the required value of electric field.

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In a cell, the amount nutrition coming in equals the amount of waste going out. This is an example of _____.
babymother [125]
The answer is B) <span>equilibrium
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5 0
3 years ago
A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a
UNO [17]

Answer:

P_{avg} = 6.283*10^{-9} \ W

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})

where;

E= \frac{d \phi}{dt}

E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t

E_{rms} =   \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}

Replacing our values into above equation; we have:

E_{rms} =   \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}

E_{rms} =   \frac {8.98909588*10^{-4} }{\sqrt{2}}

E_{rms} =   6.356*10^{-4} \ V

Then the P_{avg is calculated as:

P_{avg} = \frac{E^2}{R}

P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}

P_{avg} = 6.283*10^{-9} \ W

6 0
3 years ago
Which would fall with a greater acceleration in a vacuum-a leaf or a stone?
hichkok12 [17]

They would fall with the same acceleration

4 0
3 years ago
Read 2 more answers
A container in the shape of a cube 10.0 cm on each edge contains air (with equivalent molar mass 28.9 g/mol) at atmospheric pres
Vikentia [17]

Answer:

a) m = 1.174 grams

b) F_g = 0.01151 N

c) F_c = 1013 N

Explanation:

Given:

- The length of a cube L = 10.0 cm

- The molar mass of air M = 28.9 g/mol

- Pressure of air P = 101.3 KPa

- Temperature of air T = 300 K

- Universal Gas constant R = 8.314 J/kgK

Find:

(a) the mass of the gas

(b) the gravitational force exerted on it

(c) the force it exerts on each face of the cube

(d) Why does such a small sample exert such a great force? (6%)

Solution:

- Compute the volume of the cube:

                               V = L^3  = 0.1^3 = 0.001 m^3

- Use Ideal gas law equation and compute number of moles of air n:

                               P*V = n*R*T

                                n = P*V / R*T

                                n = 101.3*10^3 * 0.001 / 8.314*300

                                n = 0.04061 moles

- Compute the mass of the gas:

                                m = n*M

                                m = 0.04061*28.9

                                m = 1.174 grams

- The gravitational force exerted on the mass of gas is due to its weight:

                                F_g = m*g

                                F_g = 1.174*9.81*10^-3

                               F_g = 0.01151 N

- The force exerted on each face of cube is due its surface area:

                                F_c = P*A

                                F_c = (101.3*10^3)*(0.1)^2

                                F_c = 1013 N

- The molecules of a gas have high kinetic energy; hence, high momentum. When they collide with the walls they transfer momentum per unit time as force. Higher the velocity of the particles higher the momentum higher the force exerted.

4 0
3 years ago
A satellite of mass m completes one revolution around the earth at a constant speed s and radius r.
MAXImum [283]

Answer:

d. The magnitude of the work done by the earth on the satellite is non zero

Explanation:

The work done is equal to the product of the force and the distance moved in the direction of the force, the force and the distance act perpendicular to one another, therefore no work is done in the circular motion of the movement of the earth.

8 0
3 years ago
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