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KengaRu [80]
3 years ago
5

The area of the bar over r = 2 is 0.234. what is the area of the bar over r = 4?

Physics
1 answer:
natita [175]3 years ago
4 0
The area of the bar over r=4 is 0.468
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What current flow (I) is associated with an input voltage of 5.0V and resistors R1 = 1.5 kiloohms and R2 = 0.5 kiloohms? Calcula
Y_Kistochka [10]

Answer:

current in series is 2.50 mA

current in parallel is 13.51 mA

Explanation:

given data

voltage = 5 V

resistors R1 = 1.5 kilo ohms

resistors R2 = 0.5 kilo ohms

to given data

current flow

solution

current flow in series is express as here

current = voltage / resistor    .................1

put here all value  in equation 1

current = 5 / (1.5 + 0.5)

current = 5 / 2.0

so current = 2.50 mA

and

current flow in parallel is express as

current = voltage / resistor   ....................2

put here all value in equation  2

current = 5 / (1/ (1/1.5 + 1/0.5))

current = 5 / 0.37

so current = 13.31 mA

5 0
3 years ago
Please help me!
Kay [80]
Runoff because the mud is a liquid and moves on an amount of water that is in it like with quicksand
5 0
3 years ago
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If the absolute pressure of a gas is 550.280 kPa, its gage pressure is A. 101.325 kPa. B. 651.605 kPa. C. 448.955 kPa. D. 277.28
guajiro [1.7K]

Answer:

Option C is the correct answer.

Explanation:

Absolute  pressure is sum of gauge pressure and atmospheric pressure.

That is

               P_{abs}=P_{gauge}+P_{atm}

We have

          P_{abs}=550.280 kPa\\\\P_{atm}=1atm=101325Pa=101.325kPa

Substituting

         P_{abs}=P_{gauge}+P_{atm}\\\\550.280=P_{gauge}+101.325\\\\P_{gauge}=448.955kPa

Option C is the correct answer.

7 0
3 years ago
Light of wavelength 560 nm passes through a slit of width 0. 170 mm. (a) the width of the central maximum on a screen is 8. 00 m
faltersainse [42]

The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

<h3>How to find the distance between slit and the screen?</h3>
  • It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.
  • We have the expression for slit width w as,

                           w=\frac{2*wavelength*d}{a}

where, d is the distance between slit and the screen, and a is the slit width.

  • Thus, distance between slit and the screen is,

                           d=\frac{w*a}{2*wavelength} =\frac{8*10^{-3}*0.17*10^{-3}}{560*10^{-9}*2} \\\\d=1.214m

Thus, we can conclude that, the distance between slit and the screen is 1.214m.

Learn more about the width of the central maximum here:

brainly.com/question/13088191

#SPJ4

3 0
2 years ago
35 miles North 6 feet down 136 MB 85 km SE
pantera1 [17]
The answer is 133 Se
4 0
3 years ago
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