Answer:
(a) 11.3 T
(b) 6.09 T
Explanation:
Current, I = 14 kA = 14000 A
number of turns, N = 900
inner radius, r = 0.7 m
outer radius, R = 1.3 m
The magnetic field due to a circular coil is given by
![B = \frac{\mu o}{4\pi}\times \frac{2 N\pi I}{R}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%5Cmu%20o%7D%7B4%5Cpi%7D%5Ctimes%20%5Cfrac%7B2%20N%5Cpi%20I%7D%7BR%7D)
(a) The magnetic field due to the inner radius is
![B = 10^{-7}\times \frac{2\times 900\times 3.14\times 14000}{0.7}\\\\B = 11.3 T](https://tex.z-dn.net/?f=B%20%3D%2010%5E%7B-7%7D%5Ctimes%20%5Cfrac%7B2%5Ctimes%20900%5Ctimes%203.14%5Ctimes%2014000%7D%7B0.7%7D%5C%5C%5C%5CB%20%3D%2011.3%20T)
(b) The magnetic field due to the outer radius is
![B = 10^{-7}\times \frac{2\times 900\times 3.14\times 14000}{1.3}\\\\B = 6.09 T](https://tex.z-dn.net/?f=B%20%3D%2010%5E%7B-7%7D%5Ctimes%20%5Cfrac%7B2%5Ctimes%20900%5Ctimes%203.14%5Ctimes%2014000%7D%7B1.3%7D%5C%5C%5C%5CB%20%3D%206.09%20T)
The ground is very large an small amount of electric charge wont affect it
Answer:
350 ft/s²
Explanation:
First, convert mph to ft/s.
58 mi/hr × (5280 ft/mi) × (1 hr / 3600 s) = 85.1 ft/s
Given:
v₀ = 85.1 ft/s
v = 0 ft/s
t = 0.24 s
Find: a
v = at + v₀
a = (v − v₀) / t
a = (0 ft/s − 85.1 ft/s) / 0.24 s
a = -354 ft/s²
Rounded to two significant figures, the magnitude of the acceleration is 350 ft/s².
Answer: D
Neither A nor B
Explanation:
In order to check the clearances for rod and main bearings, you need a set of micrometers and a dial-bore gauge
Measuring the inside diameter of a main or rod bearing will require a dial bore gauge. The best ones to use are accurate down to 0.0001-inch.
So, both technician A and B are incorrect