Answer:
![\mathbf{ current(I) =1766.67 \ A}](https://tex.z-dn.net/?f=%5Cmathbf%7B%20current%28I%29%20%3D1766.67%20%5C%20A%7D)
Explanation:
Given that:
The air resistance and friction = 700 N
The gravity caused force = 716 × 9.8 = 7016.8
Total force = (7016.8 + 700) N
Total force = 7716.8 N
∴
![13 \times current(I) \times 0.84 = \dfrac{7716.8 \times 300}{2 \times 60}](https://tex.z-dn.net/?f=13%20%5Ctimes%20%20current%28I%29%20%5Ctimes%200.84%20%3D%20%5Cdfrac%7B7716.8%20%5Ctimes%20300%7D%7B2%20%5Ctimes%2060%7D)
![current(I) \times 10.92= 19292](https://tex.z-dn.net/?f=current%28I%29%20%5Ctimes%2010.92%3D%2019292)
![current(I) = \dfrac{19292}{10.92}](https://tex.z-dn.net/?f=current%28I%29%20%3D%20%5Cdfrac%7B19292%7D%7B10.92%7D)
![\mathbf{ current(I) =1766.67 \ A}](https://tex.z-dn.net/?f=%5Cmathbf%7B%20current%28I%29%20%3D1766.67%20%5C%20A%7D)
c.magnetic induction hope this helps
Answer:
7.05 Volts/m
Explanation:
L = length of the Nichrome wire = 44 cm = 0.44 m
V = Potential difference across the end of the wire = battery voltage = 3.1 Volts
E = magnitude of electric field inside the wire
Magnitude of electric field inside the wire is given as
![E = \frac{V}{L}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7BV%7D%7BL%7D)
Inserting the values
![E = \frac{3.1}{0.44}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B3.1%7D%7B0.44%7D)
E = 7.05 Volts/m
Given data:
* The extension of the steel wire is 0.3 mm.
* The length of the wire is 4 m.
* The area of cross section of wire is,
![A=2\times10^{-6}m^2](https://tex.z-dn.net/?f=A%3D2%5Ctimes10%5E%7B-6%7Dm%5E2)
* The young modulus of the steel is,
![Y=2.1\times10^{11}\text{ Pa}](https://tex.z-dn.net/?f=Y%3D2.1%5Ctimes10%5E%7B11%7D%5Ctext%7B%20Pa%7D)
Solution:
The young modulus of the steel in terms of the force and extension is,
![Y=\frac{F\times l}{A\times dl}](https://tex.z-dn.net/?f=Y%3D%5Cfrac%7BF%5Ctimes%20l%7D%7BA%5Ctimes%20dl%7D)
where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,
Substituting the known values,
![\begin{gathered} 2.1\times10^{11}=\frac{F\times4}{2\times10^{-6}\times0.3\times10^{-3}} \\ F=0.315\times10^2\text{ N} \\ F=31.5\text{ N} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%202.1%5Ctimes10%5E%7B11%7D%3D%5Cfrac%7BF%5Ctimes4%7D%7B2%5Ctimes10%5E%7B-6%7D%5Ctimes0.3%5Ctimes10%5E%7B-3%7D%7D%20%5C%5C%20F%3D0.315%5Ctimes10%5E2%5Ctext%7B%20N%7D%20%5C%5C%20F%3D31.5%5Ctext%7B%20N%7D%20%5Cend%7Bgathered%7D)
Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.