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maria [59]
3 years ago
10

What is kirchoff s law???

Physics
2 answers:
kolbaska11 [484]3 years ago
8 0
There are two laws named for Kirchhoff.  The both concern electrical circuits.
Here they are in my own words:

1).  The sum of the voltage drops around any closed loop in a circuit is zero.

2).  The sum of the currents at any single point in a circuit is zero.
Luda [366]3 years ago
7 0
<em>there are two
<u>1 Kirchhoff's current law:
</u>it states that the sum of all currents flowing in all conductors that meet at single point is zero..
<u>2 Kirchhoff's voltage law:
</u>
it states that the algebraic sum of all voltages around any closed network is zero...

</em>
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3 years ago
The water behind Grand Coulee Dam is 1000 m wide and 200 m deep. Find the hydrostatic force on the back of the dam. (Hint: the t
Vika [28.1K]

Answer:

The hydro static force on the back of the dam is 1.96\times10^{11}\ N

Explanation:

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Width b= 1000 m

Depth d= 200 m

We need to calculate the average pressure

Using formula of  average pressure

P_{avg}=\rho\times g\times d_{avg}

Put the value into the formula

P_{avg}=1000\times9.8\times100

P_{avg}=980000\ Pa

We need to calculate  the hydro static force on the back of the dam

Using formula of force

F = P_{avg}\times A

Put the value into the formula

F = 980000\times1000\times200

F=1.96\times10^{11}\ N

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7 0
3 years ago
You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th
PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is 6.378\times 10^{10}N/m^2

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}

where,

Y = Young's Modulus

F = force exerted by the weight  = m\times g

m = mass of the ball = 10 kg

g = acceleration due to gravity = 9.81m/s^2

l = length of wire  = 2.6 m

A = area of cross section  = \pi r^2

r = radius of the wire = \frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m      (Conversion factor:  1 m = 1000 mm)

\Delta l = change in length  = 1.99 mm = 1.99\times 10^{-3}m

Putting values in above equation, we get:

Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2

Hence, the Young's modulus for the wire is 6.378\times 10^{10}N/m^2

3 0
4 years ago
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