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klasskru [66]
3 years ago
10

A piston above a liquid in a closed container has an area of 1m2. The piston carries a load of 350 kg. What will be the external

pressure on the upper surface of the liquid?
A. 27.3 kPa
B. 3.43 kPa
C. 4.90 kPa
D. 68.6 kPa
Physics
1 answer:
RoseWind [281]3 years ago
6 0
<h3><u>Answer;</u></h3>

B. 3.43 kPa

<h3><u>Explanation</u>;</h3>

Pressure is given by the formula;

Pressure = Force / Area

Force is the product of mass and acceleration due to gravity. 

 Thus;  Force = mass x acceleration due to gravity

Substituting;

               Force = (350 kg) x (9.8 m/s²)

               Force = 3430 N

Then, using the first equation;

                Pressure = 3430 N / 1m² or Pa

<h3>                             <u>= 3.43 kPa</u></h3>

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Calculate the Fermi energy and the conductivity at room temperature for germanium containing 5×10^16 arsenic atoms per cubic cen
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Answer:

The Fermi energy is 0.568 eV and the conductivity at room temperature is 31.24 (Ωcm)⁻¹

Explanation:

Data given:

Nd = doping concentration = 5x10⁶/cm³

According the mass action law, the hole concentration is:

P_{0} =\frac{n_{i}^{2}  }{n_{0} } ,n_{i}=2x10^{13} /cm^{3}

E_{f} =E_{i} +KTln(\frac{n_{D} }{n_{i} } ),K=8.61x10^{-19} ,KT=0.026eV

E_{i} =\frac{E_{0} }{2} =\frac{0.63}{2} =0.315eV\\E_{f}=0.315+0.203=0.568eV

The conductivity of n-type of semi-conductor is equal to:

The mobility of Germanium is 3900 cm²/Vs

σ = 1.6x10⁻¹⁹ * 5x10¹⁶ * 3900 = 31.24 (Ωcm)⁻¹

6 0
3 years ago
An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch
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Answer:

a) \sigma_{\rm in} = -2.18~{\rm \mu C/m^2}

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Explanation:

Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:

Q_{\rm in} = \sigma A_{\rm in} = \sigma(2\pi r_1 h) = (-0.35)(2\pi (0.08) h) = -0.175h~{\rm \mu C}

Q_{\rm out} = \sigma A_{\rm out} = \sigma(2\pi r_2 h) = (-0.35)(2\pi (0.1) h) = -0.22h~{\rm \mu C}

Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.

When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.

Q_{\rm wire} = \lambda h = 1.1h~{\rm \mu C}

a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:

\sigma_2 = \frac{Q_{\rm in}}{2\pi r_1h} = \frac{-1.1h}{2\pi r_1 h} = \frac{-1.1}{2\pi(0.08)} = -2.18~{\rm \mu C/m^2}

b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.

The new surface charge density can be calculated as follows:

\sigma_{\rm out}= \frac{Q_{\rm out}}{2\pi r_2h} = \frac{0.705h}{2\pi r_2 h} = \frac{0.705}{2\pi(0.1)} = 1.12~{\rm \mu C/m^2}

c) The electric field outside the cylinder can be found by Gauss' Law:

\int{\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}

We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.

E(2\pi r h) = \frac{Q_{\rm enc}}{\epsilon_0}\\E2\pi rh = \frac{\sigma 2\pi (r_1 + r_2)h}{\epsilon_0}\\E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}

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Male turkeys work to attract female turkeys, but only the dominant turkeys of the group breed. The less dominant turkeys are a g
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So, all Turkeys work to attract female Turkeys, but only dominant Turkeys benefit from it, and the less dominant Turkeys don't benefit at all from it. That means that they are exhibiting altruism:selfless behavior.

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4 years ago
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If the archerfish spits its water 60 degrees from the horizontal aiming at an insect 1.4 m above
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Answer:

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Take up to be positive.  Given (in the y direction):

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vᵧ = 0 m/s

g = -10 m/s²

Find: v₀ᵧ

vᵧ² = v₀ᵧ² + 2aΔy

(0 m/s)² = v₀ᵧ² + 2 (-10 m/s²) (1.4 m)

v₀ᵧ = 5.29 m/s

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