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Jet001 [13]
3 years ago
15

A shunt resistance of 8 Ω is connected to a galvanometer of resistance of 100 Ω. The current in the galvanometer is 0.42 A. What

is the current in the circuit?
Physics
1 answer:
GuDViN [60]3 years ago
3 0
I have no idea sorry hm 111111
You might be interested in
2 Points
Mademuasel [1]

According to Newton's Second Law of Motion :

The Force acting on an Object is equal to Product of Mass of the Object and Acceleration produced due to the Force.

:\implies  Force acting = Mass of the Object × Acceleration

Given : Force = 50 newton and Mass of the Object = 10 kg

Substituting the respective values in the Formula, we get :

:\implies  50 N = 10 kg × Acceleration

:\implies \mathsf{Acceleration = \dfrac{50\;N}{10\;kg}}

:\implies Acceleration of the Object = 5 m/s²

4 0
3 years ago
Nitrous oxide is the chemical name for “laughing gas.” What are the names and the number of atoms in a molecule of nitrous oxide
Vesnalui [34]
2 nitrogen molecules and 1 oxygen molecule

6 0
3 years ago
Sunlight reflects from a concave piece of broken glass, converging to a point 34 cm from the glass. what is the radius of curvat
wolverine [178]
The rays of light coming from the Sun are parallel to each other, so when they are reflected by the concave piece of glass (which acts as a concave mirror) they converge into the focus of the mirror, which is
f=34 cm
The radius of curvature of a concave mirror is twice its focal length, so in this case it is:
r=2f = 2 \cdot 34 cm=78 cm

6 0
3 years ago
Can somebody help me please
Eddi Din [679]
51 inches.

This is because a stem-plot is formatted as so:

If it’s 5 on the left side of the line, anything on the right is the ones place making possible numbers 51, 53, 56, etc.

Hope this helps!
3 0
3 years ago
A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 25.0 cm long
joja [24]

Answer:

5.65487\times 10^{-8}\ Wb

1.17\times 10^{-5}\ H

-0.020475 V

Explanation:

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

N_1 = Number of turns of coil  = 25

N_2 = Number of turns of coil 2 = 300

\frac{di_2}{dt} = Rate of current increased = 1.75\times 10^3\ A/s

d = Diameter = 2 cm

r = Radius = \frac{d}{2}=\frac{2}{2}=1\ cm

A = Area = \pi r^2

Magnetic field in the solenoid is given by

B=\mu_0\frac{N_2}{l}I\\\Rightarrow B=4\pi\times 10^{-7}\frac{300}{0.25}\times 0.12\\\Rightarrow B=0.00018\ T

Magnetic flux is given by

\phi=BA\\\Rightarrow \phi=0.00018\times \pi\times 0.01^2\\\Rightarrow \phi=5.65487\times 10^{-8}\ Wb

The average magnetic flux through each turn of the inner solenoid is 5.65487\times 10^{-8}\ Wb

Mutual inductance is given by

L=\frac{N_1\phi}{i_1}\\\Rightarrow L=\frac{25\times 5.65487\times 10^{-8}}{0.12}\\\Rightarrow L=1.17\times 10^{-5}\ H

The mutual inductance of the two solenoids is 1.17\times 10^{-5}\ H

Induced emf is given by

V=-L\frac{di_2}{dt}\\\Rightarrow V=-1.17\times 10^{-5}\times 1.75\times 10^3\\\Rightarrow V=-0.020475\ V

The emf induced in the outer solenoid by the changing current in the inner solenoid is -0.020475 V

5 0
3 years ago
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