1.
Answer:
Part a)

Part b)

Explanation:
Part a)
Length of the rod is 1.60 m
diameter = 0.550 cm
now if the current in the ammeter is given as

V = 17.0 volts
now we will have


R = 0.91 ohm
now we know that



Part b)
Now at higher temperature we have


R = 0.98 ohm
now we know that



so we will have



2.
Answer:
Part a)

Part b)

Explanation:
Part a)
As we know that current density is defined as

now we have

Now we have


so we will have

Part b)
now we have

so we have


so we have


Answer:
Answered
Explanation:
Part A
According to Faraday's law the induced emf in coil is equal to negative of its rate of change of magnetic flux time the number of turns in the coil.
= 
When an emf generated by a change of magnetic flux, produced current of whose magnetic field opposes the change which produces it.
By the above equation the correct options are 1,2 and 4
Part B
Large signals of frequency of 60Hz are measured by osciloscope.
Hence the correct option is part 1.
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Answer:
yeah righto the correct answer mate would be D) :)
The evidence of this research is published in the scientific journal Nature communication.
<u>Explanation:</u>
Our solar system shaped about 4.5 billion years prior from a thick haze of interstellar gas and residue. The cloud crumbled, potentially due to the shock wave of a close by detonating star, called a supernova. At the point when this residue cloud crumbled, it framed a sun powered cloud—a turning, whirling plate of material.
The research is distributed in the latest issue of journal Nature Communications. About 4.6 billion years prior, a haze of gas and residue that in the end framed our nearby planetary group was upset. The following gravitational breakdown framed the proto-Sun with an encompassing plate where the planets were conceived.