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2 years ago

What is the locating position of the land field?

1 answer:

4 0

Any point on earth can be located by specifying its latitude and longitude, including Washington, DC, which is pictured here. Lines of latitude and longitude form an imaginary global grid system, shown in Fig. 1.17. Any point on the globe can be located exactly by specifying its latitude and longitude.

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A well penetrates an unconfined aquifer. Prior to pumping, the water level (head) is 25 meters. After a long period of pumping a

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3 years ago

Provide two programming examples in which multithreading provides better performance than a single-threaded solution. Provide on

Kobotan [32]

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Explanation:

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2 years ago

the voltage across a 5mH inductor is 5[1-exp(-0.5t)]V. Calculate the current through the inductor and the energy stored in the i

Serggg [28]

Given Information:

Inductance = L = 5 mH = 0.005 H

Time = t = 2 seconds

Required Information:

Current at t = 2 seconds = i(t) = ?

Energy at t = 2 seconds = W = ?

Answer:

Current at t = 2 seconds = i(t) = 735.75 A

Energy at t = 2 seconds = W = 1353.32 J

Explanation:

The voltage across an inductor is given as

$V(t) = 5(1-e^{-0.5t})$

The current flowing through the inductor is given by

$i(t) = \frac{1}{L} \int_0^t \mathrm{V(t)}\,\mathrm{d}t \,+ i(0)$

Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.

$i(t) = \frac{1}{0.005} \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \,+ 0\\\\i(t) = 200 \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \\\\i(t) = 200 \: [ {5\: (t + \frac{e^{-0.5t}}{0.5})]_0^t \\i(t) = 200\times5\: \: [ { (t + 2e^{-0.5t} + 2 )] \\$

$i(t) = 1000t +2000e^{-0.5t} -2000\\$

So the current at t = 2 seconds is

$i(t) = 1000(2) +2000e^{-0.5(2)} -2000\\\\i(t) = 735.75 \: A$

The energy stored in the inductor at t = 2 seconds is

$W = \frac{1}{2}Li(t)^{2}\\\\W = \frac{1}{2}0.005(735.75)^{2}\\\\W = 1353.32 \:J$

4 0

2 years ago

In Josiah Johnson Hawes and Albert Sands Southworth, Early Operation under Ether, Massachusetts General Hospital the elevated vi

Ronch [10]

C and A I think cause I don’t really remember this I done before it his to be C and A

4 0

3 years ago

A square loop of wire surrounds a solenoid. The side of the square is 0.1 m, while the radius of the solenoid is 0.025 m. The sq

Semmy [17]

Answer:

I=9.6×e^{-8} A

Explanation:

The magnetic field inside the solenoid.

B=I*500*muy0/0.3=2.1×e ^-3×I.

so the total flux go through the square loop.

B×π×r^2=I×2.1×e^-3π×0.025^2

=4.11×e^-6×I

we have that

(flux)'= -U

so differentiating flux we get

so the inducted emf in the loop.

U=4.11×e^{-6}×dI/dt=4.11×e^-6×0.7=2.9×e^-6 (V)

so, I=2.9×e^{-6}÷30

I=9.6×e^{-8} A

6 0

3 years ago

What is the locating position of the land field?​

What is the locating position of the land field?