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butalik [34]
3 years ago
5

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and the bottom of a b

uikding are 730 and 755 mmHg, respectively, determine the height of the building. Assume an average air density of 1.25 kg/m^3. Provide with a simple sketch and graph it (hydrostatic pressure).
Physics
1 answer:
erma4kov [3.2K]3 years ago
7 0

Answer:

h = 269.6 m

Explanation:

Pressure at the bottom of the building and at the top of the building must be related as

P_{top} = P_{bottom} - \rho g h

P_{top} = 730 mm Hg

P_{bottom} = 755 mm Hg

now we will have

(755 \times 10^{-3})(13.6 \times 10^3)(9.81) = P_{bottom}

P_{bottom} = 1.007 \times 10^5 Pa

P_{top} = (730\times 10^{-3})(13.6 \times 10^3)(9.81)

P_{top} = 0.974 \times 10^5

now we have

(1.007 - 0.974)\times 10^5 = 1.25 (9.81) h

h = 269.6 m

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Answer:

Explanation:

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The electric field in the cylinder is given by

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Here the area of cylinder is given by= 2*3.14*r*L

While for the outside, the area= 2*3.14*R*L

Equating both, we get

E= бR/rε₀

Now,

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V₀=V-бR/ε₀ln (r/R)

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olga2289 [7]

The first part of the question is not complete and it is;

The voltage across the terminals of a 250 nF capacitor is 50 V, A1e^(-4000t) + (A2)te^(-4000t) V, t0, What is the initial energy stored in the capacitor? Express your answer to three significant figures and include the appropriate units. t

Answer:

A) initial energy = 0.3125 mJ

B) A1 = 50 and A2 = 1,800,000

C) Capacitor Current is given by the expression;

I = e^(-4000t)[0.95 - 1800t]

Explanation:

A) In capacitors, Energy stored is given as;

U = (1/2)Cv²

Where C is capacitance and v is voltage.

So initial kinetic energy;

U(0) = (1/2)C(vo)²

From the question, C = 250 nF and v = 50V

So, U(0) = (1/2)(250 x 10^(-9))(50²) = 0.3125 x 10^(-3)J = 0.3125 mJ

B) from the question, we know that;

A1e^(-4000t) + (A2)te^(-4000t)

So, v(0) = A1e^(0) + A2(0)e^(0)

v(0) = 50

Thus;

50 = A1

Now for A2; let's differentiate the equation A1e^(-4000t) + (A2)te^(-4000t) ;

And so;

dv/dt = -4000A1e^(-4000t) + A2[e^(-4000t) - 4000e^(-4000t)

Simplifying this, we obtain;

dv/dt = e^(-4000t)[-4000A1 + A2 - 4000A2]

Current (I) = C(dv/dt)

I = (250 x 10^(-9))e^(-4000t)[-4000A1 + A2 - 4000tA2]

Thus, Initial current (Io) is;

Io = (250 x 10^(-9))[e^(0)[-4000A1 + A2]]

We know that Io = 400mA from the question or 0.4 A

Thus;

0.4 = (250 x 10^(-9))[-4000A1 + A2]

0.4 = 0.001A1 - (250 x 10^(-9)A2)

Substituting the value of A1 = 50V;

0.4 = 0.001(50) - (250 x 10^(-9)A2)

0.4 = 0.05 - (250 x 10^(-9)A2)

Thus, making A2 the subject, we obtain;

(0.4 + 0.05)/(250 x 10^(-9))= A2

A2 = 1,800,000

C) We have derived that ;

I = (250 x 10^(-9))e^(-4000t)[-4000A1 + A2 - 4000tA2]

So putting values of A1 = 50 and A2 = 1,800,000 we obtain;

I = (250 x 10^(-9))e^(-4000t)[(-4000 x 50) + 1,800,000 - 4000(1,800,000)t]

I = e^(-4000t)[0.05 + 0.45 - 1800t]

I = e^(-4000t)[0.95 - 1800t]

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