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kodGreya [7K]
3 years ago
8

A power plant produces 1000 MW to suply a city 40Km away.Current flows from the power plant on a single wire of resistance0.050

Ω/Km, through the city, and returns via the ground,assumed to have negligible resistance. At the power plant thevoltage between the wire and the ground is 115kV. a) What is thecurrent in the wire?. b) What fraction of the power is lostin transmission?
Physics
1 answer:
Westkost [7]3 years ago
8 0

Answer:

The current in wire resistance 2Ω

a). 8696 A

b). fraction power 15.1% a 115kV

Explanation:

Resistance

R=0.05Ω/Km*40km

R=2Ω

P=1000 MW

a).

P=V*I\\I=\frac{P}{V}=\frac{1000x10^{6}W}{115x10^{3}k }  =8696.65A

Using law ohm

b).

V=I*R\\I=\frac{V}{R}

P=I*I*R\\P=I^{2} *R\\P=8696.65^{2}*2\\P=151.228 x10^{6}  W

e=\frac{151.228x10^{6} }{1000x10^{6} }*100= 15.12%

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kap26 [50]

Answer:

The  frequencies are (f, f_1) =  (6615.4 \ Hz , 19846.2\ Hz)

Explanation:

From the question we are told that

  The  length of the ear canal is  l = 1.3 \ cm  =\frac{1.3}{100}  =  0.013 \ m

   The  speed of sound is assumed to be  v_s  =  344 \ m/s

Now  taking look at a typical  ear canal  we see that we assume it is  a  closed pipe

   Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

            f = \frac{v_s}{4 * l }

 substituting values  

          f = \frac{344}{4 * 0.013 }

         f = 6615.4 \ Hz

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

        f_1 =  \frac{3v_s}{4 * l}

 substituting values  

       f_1 =  \frac{3 *  344}{4 * 0.013}

       f_1 =   19846.2 \ Hz

Given that sound would be loudest in the pipe at the frequency, it implies that the child  will have an increased audible sensitivity at this  frequencies

6 0
3 years ago
The bohr shift on the oxygen-hemoglobin dissociation curve is produced by changes in
Alisiya [41]

Answer:

THE BOHR SHIFT ON THE OXYGEN-HEMOGLOBIN DISSOCIATION CURVE IS PRODUCED BY CHANGES IN THE CONCENTRATION OF CARBON IV OXIDE.

Explanation:

The oxygen-hemoglobin dissociation curve shows the relationship between the saturated hemoglobin concentration and oxygen. It shows how the blood hold on to and releases oxygen. The Bohr shift can occur as a result of changes in concentration of carbon iv oxide and other factors such as acidity or pH, 2,3-bisphosphoglycerate, exercise, also temperature of the body. These factors contributes to the right or left shift on the curve. Carbon iv oxide prevents the binding of oxygen to the hemoglobin. The is because hemoglobin has the same binding site for both oxygen and carbon iv oxide. Carbon iv oxide increase also leads to a change in the pH of the blood through the formation of bicarbonate ion. Bicarbonate ion formation causes reduced acidity and therefore lead a shift in the dissociation curve for more of the carbon iv oxide to be excreted as hemoglobin's affinity for oxygen reduces. And when the concentration of carbon iv oxide is low in the plasma, acidity increases and this provides more affinity for oxygen by the hemoglobin.

7 0
3 years ago
Sally has 5 bracelets and decides to wear 3 of them to school. How many ways can she choose 3 bracelets if the order she wears t
Montano1993 [528]

Answer:

60

first choice = 1/5

second choice = 1/4

third choice = 1/3

5*4*3 = 60   the number of choices

7 0
2 years ago
a coil of 40 henry inductance is connected in series with aresistance of 8ohm and the combination is joined to the terminals of
Andreyy89

Answer: 5 seconds

Explanation:

Given the following :

Inductance (L) = 40 henry

Resistance = 8 ohms

The circuit given above is a Resistor - Inductor (RL) circuit network. The time constant of an RL circuit is the ratio of the circuit Inductance (L) and Resistance (R). Time constant is measured in seconds.

THAT IS;

Time constant = L / R

THEREFORE ;

Time constant = 40 / 8

Time constant = 5 seconds

7 0
3 years ago
A physics book is thrown horizontally at a velocity of 5.0 m/s from the top of a cliff 78.4 m high. How long does the book take
xenn [34]

The book's vertical position in the air is

y=78.4\,\mathrm m-\dfrac12gt^2

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0=78.4\,\mathrm m-\dfrac12gt^2\implies t=4.00\,\mathrm s

So it takes the book 4 seconds to reach the bottom. The given initial velocity is irrelevant since it only has a horizontal component; vertically, the book is starting from rest.

3 0
3 years ago
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