Answer:
Current, I = 1000 A
Explanation:
It is given that,
Length of the copper wire, l = 7300 m
Resistance of copper line, R = 10 ohms
Magnetic field, B = 0.1 T
Resistivity, 
We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :
r = 0.00199 m
or
The magnetic field on a current carrying wire is given by :
I = 1000 A
So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.
Answer:
A star with 15 solar masses is too big to be a main-sequence star.
Answer:
270 mi/h
Explanation:
Given that,
To the south,
v₁ = 300 mi/h, t₁ = 2 h
We can find distance, d₁
To the north,
v₂ = 250 mi/h, d₂ = 750 miles
We can find time, t₂
Now,
Average speed = total distance/total time
Hence, the average speed for the trip is 270 mi/h.
Answer:
Explanation:
Using the magnification formula.
Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)
M = v/u = H1/H2
v/u = H1/H2...1
3) Given the radius of curvature of the concave lens R = 20cm
Focal length F = R/2
f = 20/2
f = 10cm
Object distance u = 5cm
Object height H2= 5cm
To get the image distance v, we will use the mirror formula
1/f = 1/u+1/v
1/v = 1/10-1/5
1/v = (1-2)/10
1/v =-1/10
v = -10cm
Using the magnification formula
(10)/5 = H1/5
10 = H1
H1 = 10cm
Image height of the peg is 10cm
4) If u = 15cm
1/v = 1/f-1/u
1/v = 1/10-1/15
1/v = 3-2/30
1/v = 1/30
v = 30cm
30/15 = H1/5
15H1 = 150
H1/= 10cm
5) if u = 20cm
1/v = 1/f-1/u
1/v = 1/10-1/20
1/v = 2-1/20
1/v = 1/20
v = 20cm
20/20 = H1/5
20H1 = 100
H1 = 5cm
6) If u = 30cm
1/v = 1/f-1/u
1/v = 1/10-1/30
1/v = 3-1/30
1/v = 2/30
v = 30/2 cm
v =>15cm
15/30 = Hi/5
30H1 = 75
H1 = 75/30
H1 = 2.5cm
