Answer:
0.45 miles
Explanation:
since, car will be running on the edge of track, distance covered in one lap by car will be equal to circumference of track.\
we will use value of pi as 22/7
Circumference for any circular shape is given by 
Hence, in one lap distance will be covered
in 7 laps 
distance will be covered.
but is given that 7 laps should be equal to 10 miles
so 10 miles will be equal to 
Thus, diameter of track will be 0.45 miles to meet the requirement.
Explanation:
ω = 30 rev × (2π rad/rev) / (6 sec)
ω = 10π rad/s
ω ≈ 31.4 rad/s
Answer:
The period the field must be reduced to zero is 9.81 x 10⁻⁵ s
Explanation:
Given;
initial value of the magnetic field, B₁ = 0.276 T
number of turns of the solenoid, N = 517 turns
diameter of the solenoid, d = 10.5 cm = 0.105 m
induced emf, = 12.6 kV = 12,600 V
when the field becomes zero, then the final magnetic field value, B₂ = 0
The induced emf is given by Faraday's law;
Therefore, the period the field must be reduced to zero is 9.81 x 10⁻⁵ s
Answer:
ΔP = 20000 N s
Explanation:
To solve this problem we use the relation between momentum and moment
I = Δp
let's calculate the momentum
I = ∫F dt
if we use the average force
I = F t
I = 10000 2
I = 20000 N s
therefore with the first equation
ΔP = I = 20000 N s
image distance,di=10 cm
object distance,do=20cm
magnification, m=di/do
=10/20
=0.5
since the image is virtual, magnification is negative.
therefore m=-0.5
