Answer:
A) determine magnetic fields
For 0 ≤ r ≤ a
Magnetic field = ∅ 
For a ≤ r ≤ b
Magnetic field = ∅ 
For b ≤ r ≤ c
Magnetic field in the region = ∅ ![\frac{I}{2\pi r} [ c^2 - r^2 / c^2-b^2 ]](https://tex.z-dn.net/?f=%5Cfrac%7BI%7D%7B2%5Cpi%20r%7D%20%5B%20c%5E2%20-%20r%5E2%20%2F%20c%5E2-b%5E2%20%5D)
For r ≥ c
magnetic filed in the region = 0
B ) attached below
Explanation:
<u>A) Determine the magnetic field in the following regions</u>
i) For 0 ≤ r ≤ a
Magnetic field = ∅
attached below is the detailed solution
ii) For a ≤ r ≤ b
Magnetic field = ∅
attached below is the detailed solution
iii) For b ≤ r ≤ c
Magnetic field in the region = ∅
attached below is the detailed solution
iv) For r ≥ c
magnetic filed in the region = 0 and this is because the net current enclosed in the region = 0
Answer: W = 294 J
Explanation: Solution:
Work is expressed as the product of force and the distance of the object.
W = Fd where F = mg
W= Fd
= mg d
= 15 kg ( 9.8 m/s²) ( 2m )
= 294 J
Answer:
Explanation:
mass of baseball, m = 0.145 kg
initial velocity, u = 12 m/s upward
(a) final velocity, v = u / 2 = 6 m/s
Let the height is h.
Use third equation of motion
v² = u² - 2gh
6 x 6 = 12 x 12 - 2 x 9.8 x h
36 - 144 = - 19.6 x h
h = 5.51 m
(b) initial kinetic energy, K = 0.5 x m x u² = 0.5 x 0.145 x 12 x 12 = 10.44 J
Final kinetic energy, K' = K/2
0.5 x m x v² = 10.44 /2
0.5 x 0.145 x v² = 5.22
v = 8.5 m/s
v² = u² - 2gh
8.5 x 8.5 = 12 x 12 - 2 x 9.8 x h
72.25 - 144 = - 19.6 x h
h = 3.66 m
Below is the solution:
Heat soda=heat melon
<span>m1*cp1*(t-t1)=m2*cp2*(t2-t); cp2=cpwater </span>
<span>12*0.35*3800*(t-5)=6.5*4200*(27-t) </span>
<span>15960(t-5)=27300(27-t) </span>
<span>15960t-136500=737100-27300t </span>
<span>43260t=873600 </span>
<span>t=873600/43260 </span>
<span>t=20.19 deg celcius</span>
