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Mkey [24]
3 years ago
13

What are the precaution you take when you are carrying out experiment on kirchhoffs law​

Physics
1 answer:
sdas [7]3 years ago
4 0

Answer:

The precautions to be followed for performing kirchoff law experiment are:-

  1. check that all the connections inside the arrangement are tight and clean.
  2. check that the terminals of the resistances  are nicely attached.
  3. ensure that the instruments used in the experiment set-up are indicating zero.(before starting of the experiment)

Explanation:

These are the basic precautions that must be kept in mind before executing kirchoff law experiment.

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Total magnetic field at the point P midway between the wires is Zero.

The higher wire's contribution to the magnetic field at point P is directed into the page, whereas the lower wire's contribution is directed out of the page. These two oppositely directed contributions to the magnetic field have identical magnitudes and cancel each other out since point P is equally spaced from the two wires and the currents flowing through them are of equal magnitude.

<h3></h3><h3>Define magnitude?</h3>

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5 0
1 year ago
Clouds form from _____ temperature change, which occurs when an expanding gas cools.
IrinaVladis [17]
The answer is Adiabatic
8 0
3 years ago
Why the watt is called derived unit ?​
Yuri [45]

Answer:

It is because it is a unit of Power(which is a derived quantity)

4 0
3 years ago
The drawing shows four sheets of polarizing material, each with its transmission axis oriented differently. Light that is polari
Brrunno [24]

Answer:

Explanation:

When sheet A is removed

I_B=32\cos^230=24W/m^2\\\\I_C=24 \cos^260=6W/m^2\\\\I_D=6\cos^230=4.5W/m^2

When sheet B is removed

I_A=32\cos^20=32W/m^2\\\\I_C=24 \cos^290=0W/m^2\\\\I_D=0\cos^230=0W/m^2

When sheet C is removed

I_A=32\cos^20=32W/m^2\\\\I_D=32 \cos^230=24W/m^2\\\\I_B=24\cos^290=0W/m^2

When sheet D is removed

I_A=32\cos^20=32W/m^2\\\\I_B=32\cos^230=24W/m^2\\\\I_C=24\cos^260=6W/m^2

8 0
3 years ago
A river flows with a uniform velocity vr. A person in a motorboat travels 1.22 km upstream, at which time she passes a log float
storchak [24]

Answer:

 t ’= \frac{1450}{0.6499 + 2 v_r},  v_r = 1 m/s       t ’= 547.19 s

Explanation:

This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.

By the time the boat goes up the river

        v_b - v_r = d / t

By the time the boat goes down the river

        v_b + v_r = d '/ t'

let's subtract the equations

       2 v_r = d ’/ t’ - d / t

       d ’/ t’ = 2v_r + d / t

       t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }

In the exercise they tell us

         d = 1.22 +1.45 = 2.67 km= 2.67 10³ m

         d ’= 1.45 km= 1.45 1.³ m

at time t = 69.1 min (60 s / 1min) = 4146 s

the speed of river is v_r

      t ’= \frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \  + 2 \ v_r}

      t ’= \frac{1450}{0.6499 + 2 v_r}

In order to complete the calculation, we must assume a river speed

          v_r = 1 m / s

       

let's calculate

      t ’= \frac{ 1450}{ 0.6499 + 2 \ 1}

      t ’= 547.19 s

8 0
3 years ago
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