Answer:
Length = 2.32 m
Explanation:
Let the length required be 'L'.
Given:
Resistance of the resistor (R) = 3.7 Ω
Radius of the rod (r) = 1.9 mm = 0.0019 m [1 mm = 0.001 m]
Resistivity of the material of rod (ρ) = 
First, let us find the area of the circular rod.
Area is given as:

Now, the resistance of the material is given by the formula:

Express this in terms of 'L'. This gives,

Now, plug in the given values and solve for length 'L'. This gives,

Therefore, the length of the material required to make a resistor of 3.7 Ω is 2.32 m.
Answer:
the electroscope separate by the presence of charge carriers
Explanation:
Metal bodies are characterized by having free (mobile) electrons. In the electroscope the plates are in balance; when the external metal ball is touched, a charge is introduced into the device, when the body that touched the ball is separated, an excess charge remains. This charge, being a metal, is distributed over the entire surface, giving a uniform density and an electric force of repulsion is created between the two charged sheets, which tends to separate the sheets. This force is counteracted by the tension component as the sheets are separated at a given angle, the separation reaches the point where
Fe - Tx = 0
Fe = Tx
In summary, the electroscope separate its leaves by the presence of charge carriers
Answer:
Following are the responses to these question:
Explanation:
Since the
is the current of ckt which depend on the reactance which inductor that also enables the ckt and inductor resistance
for capacities
for

When 
then
therefore,
remains at the same so, the maximum current remains the in same ckt.
Answer:
The nodes and anti nodes would reverse roles.
Explanation:
I believe it has to do with the path differences. If waves are in phase, then the path differences are such that the waves reach the screen with crests superimposing crests and troughs superimposing troughs. This happens when the periods of each wave are equal or the paths themselves differ by a whole number multiple of the wavelength (λ, 2λ, 3λ, ...).
Now make these waves out of phase. Then half of the waves will travel half a wavelength farther than the rest. So the path difference will be 0.5λ, 1.5λ, 2.5λ, ....
Answer:
divide the mass value by 1e+8