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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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Answer:

11 = 11

Explanation:

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3 years ago
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When 597 J of heat are added to a gas, it expands. Its internal energy increases by 318 J. How much work does the gas do? (Unit=
damaskus [11]

The magnitude of work done by the gas is 279 J and the sign is negative so W = -279 J as work is done by the system.

<u>Explanation:</u>

According to first law of thermodynamics, the change in internal energy of the system is equal to the sum of the heat energy added or released from the system with the work done on or by the system. If the heat energy is added to the system to perform a certain work, then the heat energy is taken as positive, while it will be negative when the heat energy is released from the system.

Similarly, in this case, the heat energy of 597 J is added to the system. So the heat energy will be positive, while the gas expansion occurs means work is done by the system.

          ΔU = Q+W

Since ΔU is the change in internal energy which is given as 318 J and the heat energy added to the system is Q = 597 J.

Then the work done by the gas = ΔU - Q = 318 J - 597 J = - 279 J.

As the work is done by the system, so it will be denoted in negative sign and the magnitude of work done by the gas is 279 J.

7 0
3 years ago
A wire of Nichrome (a nickel– chromium– iron alloy commonly used in heating elements) is 1.0 m long and 1.0 mm² in cross-se
Lyrx [107]

Answer:

The conductivity of Nichrome is 2\times 10^6\ S/m.

Explanation:

Given:

Potential difference (V) = 2.0 V

Current flowing (I) = 4.0 A

Length of wire (L) = 1.0 m

Area of cross section of wire (A) = 1.0 mm² = 1 × 10⁻⁶ m² [1 mm² = 10⁻⁶ m²]

We know, from Ohm's law, that the ratio of voltage and current is always a constant and equal to the resistance of the resistor. Therefore, the resistance of the nichrome wire is given as:

R=\frac{V}{A}=\frac{2.0}{4.0}=0.5\ \Omega

Now, resistance of the nichrome wire in terms of its resistivity, length and area of cross section is given as:

R=\rho\frac{L}{A}

Where, \rho\to resistivity\ of\ Nichrome

Now, plug in all the values given and solve for \rho. This gives,

0.5\ \Omega=\rho\frac{1.0\ m}{1\times 10^{-6}\ m^2}\\\\\rho=\frac{0.5\times 1\times 10^{-6}}{1.0}=0.5\times 10^{-6}\ \Omega-m

Now, conductivity of a material is the reciprocal of its resistivity. Therefore, the conductivity of Nichrome is given as:

\sigma=\frac{1}{\rho}=\frac{1}{0.5\times 10^{-6}}=2\times 10^6\ S/m

Conductivity is measured in Siemens per meter (S/m)

Therefore, the conductivity of Nichrome is 2\times 10^6\ S/m.

7 0
3 years ago
A student throws a 5.0 newton ball straight up. What is the net force on the ball at its maximum height
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Particle A has half the mass and eight times the kinetic energy of particle B. What is the speed ratio vA / vB?
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Answer:

Use the equation "KE=½mv²", and use some algebra. > "Particle A has two times the mass...of particle B" mA = 2mB > "Particle A has...8 situations the kinetic power of particle B" KE_A = 8(KE_B) or: ½(mA)(vA)² = 8(½(mB)(vB)²) the rest is uncomplicated algebra: in basic terms sparkling up the above equation for "vA/vB". (hint: start up by utilising dividing the two factors by utilising "(mB)(vB)²". Then make the substitution: mA/mB = 2 (from the 1st eq0.5 ma *  VA^2 = 8 * 0.5 * mb * VB^2

0.5 * 0.5*mb *  VA^2 = 8 * 0.5 * mb * VB^2

0.5 * 0.5*  VA^2 = 8 * 0.5 * VB^2

VA/VB = 4uation))

Explanation:

8 0
3 years ago
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