Answer:
1.144 A
Explanation:
given that;
the length of the wire = 2.0 mm
the diameter of the wire = 1.0 mm
the variable resistivity R = ![\rho (x) =(2.5*10^{-6})[1+(\frac{x}{1.0 \ m})^2]](https://tex.z-dn.net/?f=%5Crho%20%28x%29%20%3D%282.5%2A10%5E%7B-6%7D%29%5B1%2B%28%5Cfrac%7Bx%7D%7B1.0%20%5C%20m%7D%29%5E2%5D)
Voltage of the battery = 17.0 v
Now; the resistivity of the variable (dR) can be expressed as = 
![dR = \frac{(2.5*10^{-6})[1+(\frac{x}{1.0})^2]}{\frac{\pi}{4}(10^{-3})^2}](https://tex.z-dn.net/?f=dR%20%3D%20%5Cfrac%7B%282.5%2A10%5E%7B-6%7D%29%5B1%2B%28%5Cfrac%7Bx%7D%7B1.0%7D%29%5E2%5D%7D%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%2810%5E%7B-3%7D%29%5E2%7D)
Taking the integral of both sides;we have:
![\int\limits^R_0 dR = \int\limits^2_0 3.185 \ [1+x^2] \ dx](https://tex.z-dn.net/?f=%5Cint%5Climits%5ER_0%20%20dR%20%3D%20%5Cint%5Climits%5E2_0%203.185%20%5C%20%5B1%2Bx%5E2%5D%20%5C%20dx)
![R = 3.185 [x + \frac {x^3}{3}}]^2__0](https://tex.z-dn.net/?f=R%20%3D%203.185%20%5Bx%20%2B%20%5Cfrac%20%7Bx%5E3%7D%7B3%7D%7D%5D%5E2__0)
![R = 3.185 [2 + \frac {2^3}{3}}]](https://tex.z-dn.net/?f=R%20%3D%203.185%20%5B2%20%2B%20%5Cfrac%20%7B2%5E3%7D%7B3%7D%7D%5D)
R = 14.863 Ω
Since V = IR


I = 1.144 A
∴ the current if this wire if it is connected to the terminals of a 17.0V battery = 1.144 A
(A) 19.2 W
<u>Explanation:</u>
Given-
Voltage drop, V = 24 V
Resistor = 30Ω
Current, I = 0.8 A
Power, P = ?
We know,
P = VI
P = 24 (0.8)
P = 19.2 W
Therefore, the power conducted by the resistor is 19.2 W
To solve this problem it is necessary to apply the concepts related to Dopler's Law. Dopler describes the change in frequency of a wave in relation to that of an observer who is in motion relative to the Source of the Wave.
It can be described as

c = Propagation speed of waves in the medium
= Speed of the receiver relative to the medium
= Speed of the source relative to the medium
Frequency emited by the source
The sign depends on whether the receiver or the source approach or move away from each other.
Our values are given by,
Velocity of car
velocity of motor
Velocity of sound
Frequency emited by the source
Replacing we have that



Therefore the frequency that hear the motorcyclist is 601.7Hz
Answer:
C. Quadruple
Explanation:
¨Drivers who are talking on the phone, even on a hands-free device, are up to four times more likely to be involved in a crash.¨
I hope this helps! Have a great day!