Answer:
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Answer:a)1.11×10^-21Nm
b) 1.16×10^-3m
Explanation:see attachment
5 minutes is the minimum time
Answer:
Explanation:
The cars are moving away from the intersection at 90° from each other. The motion can be considered a right angled triangle with perpendicular and base being the speeds of the 2 cars. The hypotenuse is the linear distance between them. By Pytagoras theorem:
a) Distance=
where a and b are the speeds of 2 cars.
Distance =
=77.2t feet/second
b) Time = 2 hours 45 minutes; Convert this into seconds to get 9900 seconds. The distance formula will give distance in feet so we will divide it by 5280 to get miles
Distance= (77.2*9900)/5280 = 144.8 miles
c) 1 Miles = 5280 feet.
5280=77.2t;
Time=68.4 seconds;
The 2 cars are 1 miles apart after 68.4 seconds.
Answer:
1.117935:1
Explanation:
Since the wires are of the same material, they will have the same resistivity
.
The cross-sectional area of the of a wire is given by;
![A=\pi\frac{d^2}{4}................(1)](https://tex.z-dn.net/?f=A%3D%5Cpi%5Cfrac%7Bd%5E2%7D%7B4%7D................%281%29)
where d is the diameter of the wire.
Also, the relationship between resistance R, cross-sectional area A and length l of a wire is given as follows;
![\rho=\frac{RA}{l}..................(2)](https://tex.z-dn.net/?f=%5Crho%3D%5Cfrac%7BRA%7D%7Bl%7D..................%282%29)
Since the resistivity same for both wires, say wire 1 and wire 2, we can wreite the following;
![\frac{R_1A_1}{l_1}=\frac{R_2A_2}{l_2}..................(3)](https://tex.z-dn.net/?f=%5Cfrac%7BR_1A_1%7D%7Bl_1%7D%3D%5Cfrac%7BR_2A_2%7D%7Bl_2%7D..................%283%29)
Hence from eqaution (3), the ration of wire 1 to 2 is expressed as;
![\frac{R_1}{R_2}=\frac{l_1A_2}{l_2A_1}..................(4)](https://tex.z-dn.net/?f=%5Cfrac%7BR_1%7D%7BR_2%7D%3D%5Cfrac%7Bl_1A_2%7D%7Bl_2A_1%7D..................%284%29)
Given;
![l_1=1.35l_2](https://tex.z-dn.net/?f=l_1%3D1.35l_2)
![\frac{R_1}{R_2}=\frac{1.35l_2A_2}{l_2A_1}\\\frac{R_1}{R_2}=\frac{1.35A_2}{A_1}.............(5)](https://tex.z-dn.net/?f=%5Cfrac%7BR_1%7D%7BR_2%7D%3D%5Cfrac%7B1.35l_2A_2%7D%7Bl_2A_1%7D%5C%5C%5Cfrac%7BR_1%7D%7BR_2%7D%3D%5Cfrac%7B1.35A_2%7D%7BA_1%7D.............%285%29)
We then use equation (1) to fine the ratio of the area
to ![A_2](https://tex.z-dn.net/?f=A_2)
bearing in mind that ![d_1=0.91d_2](https://tex.z-dn.net/?f=d_1%3D0.91d_2)
This ratio gives 0.8281. Substituting this into equation (5), we get the following;
![\frac{R_1}{R_2}= 1.35*0.8281=1.117935](https://tex.z-dn.net/?f=%5Cfrac%7BR_1%7D%7BR_2%7D%3D%201.35%2A0.8281%3D1.117935)