Answer:
54.0m
Explanation:
#First we solve for
![R=p(L/A)\\\\L=RA/p](https://tex.z-dn.net/?f=R%3Dp%28L%2FA%29%5C%5C%5C%5CL%3DRA%2Fp)
let's denote the initial and final length of the rope as
respectively and given as:
![L_o=\frac{R_oA}{p}, \ \ \ \ L_f=\frac{R_fA}{p}\ \ \ \ \ .....eqtn1\\](https://tex.z-dn.net/?f=L_o%3D%5Cfrac%7BR_oA%7D%7Bp%7D%2C%20%5C%20%5C%20%5C%20%5C%20L_f%3D%5Cfrac%7BR_fA%7D%7Bp%7D%5C%20%5C%20%20%5C%20%5C%20%5C%20.....eqtn1%5C%5C)
is the initial resistance and Rf the final of the wire.
![\frac{L_f}{L_o}=\frac{R_fA/P}{R_oA/P}=R_f/R_o\\\\\\or \ L_f=(R_f/R_o)L_o\ \ \ \ \ \ ...eqtn2](https://tex.z-dn.net/?f=%5Cfrac%7BL_f%7D%7BL_o%7D%3D%5Cfrac%7BR_fA%2FP%7D%7BR_oA%2FP%7D%3DR_f%2FR_o%5C%5C%5C%5C%5C%5Cor%20%5C%20L_f%3D%28R_f%2FR_o%29L_o%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20...eqtn2)
From
the initial resistance
of the spooled wire are:
![R_o=V/I_o,\ \ \ \ \ \ R_f=V/I_f \ \ \ \ \ \ \....eqtn3\\](https://tex.z-dn.net/?f=R_o%3DV%2FI_o%2C%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20R_f%3DV%2FI_f%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C....eqtn3%5C%5C)
#Substituting eqtn 3 in 2, we get
![L_f=(V/I_f)/(V/I_o)L_o\\\\=\frac{2.7A}{3.5A}\times 70m\\\\=54m](https://tex.z-dn.net/?f=L_f%3D%28V%2FI_f%29%2F%28V%2FI_o%29L_o%5C%5C%5C%5C%3D%5Cfrac%7B2.7A%7D%7B3.5A%7D%5Ctimes%2070m%5C%5C%5C%5C%3D54m)
#the length of wire remaining on the spool is 54.0m
Given that,
Current = ∞
We know that,
Ohm's law :
Ohm's law is defined as,
The voltage of the circuit is directly proportional to the current of the circuit.
![V\propto I](https://tex.z-dn.net/?f=V%5Cpropto%20I)
Or,
The voltage of the circuit is equal to the product of current and resistance.
In mathematically,
![V=IR](https://tex.z-dn.net/?f=V%3DIR)
Where, V = voltage
I = current
R = resistance
According to ohm's law,
The current in the circuit is
![I=\dfrac{V}{R}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7BV%7D%7BR%7D)
If the current is very less then the resistance will be infinity.
If the is reach to infinity then the resistance will be very low.
Hence, The resistance becomes very low.
Answer:
A. Heat flows from an object at higher temperature to an object at lower temperature
Explanation:
The option A obeys the 2nd law of thermodynamics. The heat will flow from the object at higher temperature to the object at Lower temperature till they reach an equilibrial state.
Heat doesn’t necessarily flow from an object with higher thermal energy to an object with lower thermal energy because an object has a higher thermal energy when it’s mass is more than the other. This makes B wrong.
C is wrong because heat moves from an object with higher temperature to objects with Lower temperature regardless of the state of matter.
Answer:
You take the light from a star, planet or galaxy and pass it through a spectroscope, which is a bit like a prism letting you split the light into its component colours. "It lets you see the chemicals being absorbed or emitted by the light source. From this you can work out all sorts of things," says Watson