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xxMikexx [17]
4 years ago
7

A student measures the electric flux through a closed spherical surface of volume V to be X. She then removes the charge from in

side the spherical surface and places it in a closed cylindrical surface of volume V/2. She then claims that the flux through the cylindrical surface is 2X. Is the student right or wrong ? Give reasons to your answers
Physics
1 answer:
ELEN [110]4 years ago
4 0

Answer:

The student is wrong

Explanation:

According to Gauss' law, electric flux through a surface is independent of the shape of the surface. So, the same electric flux which flows through the spherical surface flows through the cylindrical surface.

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What determines how the plates interact at their boundaries
swat32

Answer:

Tectonic plate interactions are of three different basic types: Divergent boundaries are areas where plates move away from each other, forming either mid-oceanic ridges or rift valleys. These are also known as constructive boundaries. Convergent boundaries are areas where plates move toward each other and collide.

Explanation:

Meaning the answer to your question is depending on what type of tectonic plate interaction is occurring will depend on how the plates interact.

3 0
3 years ago
A 55-kg mountain climber, starting from rest, climbs a vertical distance of 701 m. At the top, she is again at rest. In the proc
aksik [14]

Answer:

7.75%

Explanation:

mass of climber, m = 55 kg

height, h = 701 m

Qc = 4.5 x 10^6 J (heat exhaused by the body)

Work = m x g x h

W = 55 x 9.8 x 701

W = 377839 J

W = QH - Qc

Where, QH is the heat input

QH = 377839 + 4.5 x 10^6

QH = 4877839 J

So, the efficiency

e = W / QH

e = 377839 / 4877839

e = 0.0774 = 7.75 %

Thus, the efficiency of the body is 7.75 %.

3 0
3 years ago
Your friend, who is age 15 and has a resting heart rate of 70, wishes to start a jogging program. What would you suggest as a pr
Alex787 [66]
I think the right answer for this question is : 138 because the boy is just 15 year old but he has resting heart rate of 70
6 0
4 years ago
Read 2 more answers
A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very l
artcher [175]

Answer:

at y=6.29 cm the charge of the two distribution will be equal.

Explanation:

Given:

linear charge density on the x-axis, \lambda_1=8\times 10^{-6}\ C

linear charge density of the other charge distribution, \lambda_2=-6\times 10^{-6}\ C

Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.

Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.

<u>we know, the electric field due to linear charge is given as:</u>

E=\frac{\lambda}{2\pi.r.\epsilon_0}

where:

\lambda= linear charge density

r = radial distance from the center of wire

\epsilon_0= permittivity of free space

Therefore,

E_1=E_2

\frac{\lambda_1}{2\pi.x.\epsilon_0}=\frac{\lambda_2}{2\pi.(0.11-x).\epsilon_0}

\frac{\lambda_1}{x} =\frac{\lambda_2}{0.11-x}

\frac{8\times 10^{-6}}{x} =\frac{6\times 10^{-6}}{0.11-x}

x=0.0629\ m

∴at y=6.29 cm the charge of the two distribution will be equal.

9 0
3 years ago
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and
nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}

k = 1.4

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K

Work done is given as;

W = \frac{1}{2} *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s

6 0
3 years ago
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