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harina [27]
2 years ago
7

A 25 pF parallel-plate capacitor with an air gap between the plates is connected to a 100 V battery. A Teflon slab is then inser

ted between the plates, and completely fills the gap. What is the change in the charge on the positive plate when the Teflon is inserted?
Physics
1 answer:
fgiga [73]2 years ago
6 0

Answer:

5.3 nC

Explanation:

The initial charge stored on the capacitor is given by:

Q=CV (1)

where

C=25 pF = 25\cdot 10^{-12}F is the capacitance of the capacitor

V=100 V is the potential difference across the capacitor

Substituting numbers into the equation, we have

Q=(25\cdot 10^{-12} F)(100 V)=25\cdot 10^{-10}C=2.5 nC

When the Teflon slab is inserted between the plates, the capacitance of the capacitor is increased as follows:

C'=kC

where k=2.1 is the dielectric constant of the Teflon. Since the voltage V remains constant, this means that the new charge stored by the capacitor (1) will be

Q'=C'V=kCV=kQ

and so

Q'=(2.1)(2.5 nC)=5.3 nC

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100-ft-long horizontal pipeline transporting benzene develops a leak 43 ft from the high-pressure end. The diameter of the leak
Amanda [17]

Answer:

Explanation:

The mass flow rate of benzene from the leak in the pipeline containing benzene is:

Q_m=AC_o\sqrt{2\rho g_cP_g}

Here, Q_m is the mass flow rate through the leak of the pipeline. A is the area of the hole, C_o is the discharge rate, \rho is the fluid density, g_c is the gravitational constant and P_g is the constant gauge pressure within the process unit.

The diametre of the leak (d) is 0.1 in. Convert from in to ft.

d=(0.1 in)(\frac{1ft}{12in})\\=8.33\times 10^{-3}ft

Calculate the area (A) of the hole. The area of the hole is.

A=\frac{\pi d^2}{4}

Substitute 3.14 for \pi and 8.33\times 10^{-3}ft for d and calculate A.

A=\frac{\pi d^2}{4}\\\\\frac{(3.14)(8.33\times 10^{-3})^2}{4}\\\\5.45\times 10^{-5}ft^2

The specific gravity of benzene is 0.8794. Specific gravity is the ratio of th density of a substance to the density of a reference substance.

Specific gravity of benzene = density of benzenee/denity of reference substance

Rewrite the expression in terms of density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

Take the reference substance as water. Density of water is 62.4\frac{Ib_m}{ft^3}. Calculate density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

=(0.8794)(62.4\frac{Ib_m}{ft^3})\\\\54.9\frac{Ib_m}{ft^3}

Calculate the pressure at the point of leak. The pressure is the average of the pressure of the high and low pressure end. Write the expression to calculate the average pressure.

Upstream x distance from upstream pressure end

P_g=+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

Calculate the distance from the downstream pressure end. The distance from upstream pressure end is 43 ft. Total of the pipe is 100 ft.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

The distance from upstream pressure end is 43 ft. Total length of the pipe is 100 ft. Substitute the values in the equation.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

= 100ft - 43ft = 57 ft

Substitute 50 psig for upstream, 43 ft fr distance from the upstream pressure end, 40 psig for downstream pressure, 57 ft for distance from the downstream pressure end, and 100 ft for the total length of the horizontal pipeline and calculate P_g.

Upstream x distance from upstream pressure end

P_g=+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

=\frac{(50psig\times 43ft)+(40psig \times 57ft)}{100ft}\\\\=44.3psig

Convert the pressure from psig to Ib_f/ft^2

P_g=(44.3psig)(\frac{1\frac{Ib_f}{ft^2}}{1psig})(144\frac{in^2}{ft^2})\\\\=6,379.2\frac{Ib_f}{ft^2}

The leak is like a sharp orifice. Take the value of the discharge coefficient as 0.61.

Substitute 5.45\times 10^{-5}ft^2 for A. 0.61 for C_o, 54.9\frac{Ib_m}{ft^3} for \rho, 32.17\frac{ft.Ib_m}{Ib_f.s^2} for g_c, and 6,379.2\frac{Ib_f}{ft^2} for P_g and calculate Q_m

Q_m=AC_o\sqrt{2\rho g_cP_g}\\\\=(5.45\times 10^{-5}ft^2)(0.61)\sqrt{2(54.9\frac{Ib_m}{ft^3})(32.17\frac{ft.Ib_m}{Ib_f.s^2})(6,379.2\frac{Ib_f}{ft^2})}\\\\(3.3245\times 10^{-5}ft^2)\sqrt{22,533,031.21\frac{Ib^2_m}{ft^4.s^2}}\\\\=0.158\frac{Ib_m}{s}

The mass flow rate of benzene through the leak in the pipeline is 0.158\frac{Ib_m}{s}

8 0
3 years ago
Which one of the following statements concerning kinetic energy is true? a The kinetic energy of an object always has a positive
hodyreva [135]

a The kinetic energy of an object always has a positive value.

Explanation:

The kinetic energy of an object is the energy possessed by an object due to its motion, and it can be calculated as

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

Let's now analyze each statement:

a The kinetic energy of an object always has a positive value. --> TRUE. In fact, the mass of the object is always positive, and the term v^2 is always positive as well, so the kinetic energy is always positive.

b The kinetic energy of an object is directly proportional to its speed. --> FALSE. Looking at the formula, we see that the kinetic energy is proportional to the square of the speed, K\propto v^2.

c The kinetic energy of an object is expressed in watts. --> FALSE. Watts is the units for measuring power, while the kinetic energy is measured in Joules, the units for the energy.

d The kinetic energy of an object is a quantitative measure of its inertia. --> FALSE. The inertia of an object depends only on its mass, not on its speed.

e The kinetic energy of an object is always equal to the object’s potential energy. --> FALSE. The potential energy depends on the altitude from the ground, not from the speed, so the two energies can be different.

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

8 0
2 years ago
Hoosier Manufacturing operates a production shop that is designed to have the lowest unit production cost at an output rate of 1
abruzzese [7]

Answer:

90.77%

its capacity utilization rate for the month is 90.77%

Explanation:

The capacity utilisation rate can be expressed mathematically as;

Capacity utilisation rate = capacity used/Best operating level × 100%

Given;

Total Number of production time = 205hours

Production output/capacity used = 21400 units

Best operation rate = 115units/hour

Best operation output for the month of July( at best operation level )

=115units/hour × 205 hours = 23575 units

Capacity utilisation rate = 21400/23575 × 100%

= 90.77%

3 0
2 years ago
A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe
Nonamiya [84]

Answer:

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

=  surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ /  2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ /  2ε₀

= 6.78 X 10³ N/C

4 0
2 years ago
A person's height will increase from birth until about age 25, and it may decrease starting at about age 70. This is an example
dsp73
LOL idk sorry maybe it b! Oh wait it is !
5 0
2 years ago
Read 2 more answers
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