Answer:
Explanation:
The magnetic force acting horizontally will deflect the wire by angle φ from the vertical
Let T be the tension
T cosφ = mg
Tsinφ = Magnetic force
Tsinφ = BiL , where B is magnetic field , i is current and L is length of wire
Dividing
Tanφ = BiL / mg
= .055 x 29 x .11 / .010 x 9.8
= 1.79
φ = 61° .
Tension T = mg / cosφ
= .01 x 9.8 / cos61
= .2 N .
Answer:
1408.685 KN/C
Explanation:
Given:
R = 0.45 m
σ = 175 μC/m²
P is located a distance a = 0.75 m
k = 8.99*10^9
- The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:
part a)
Electric Field strength at point P: a = 0.75 m
part b)
Since, R >> a, we can approximate a / R = 0 ,
Hence, E simplified relation becomes:
E = σ / 2*e_o
part c)
Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:
Electric Field strength due to point charge is:
E = k*δ*pi*R^2 / a^2
Since, R << a, Surface area = δ*pi
Hence,
E = (k*δ*pi/a^2)
The wires that will have the least resistance is :
C. A short thick wire
in order to get the least resistence, you need the wire to be the lowest in length and the highest in Area
hope this helps
In vacuum, going at 2.99×10^8 m/s.
The net speed due west is = distance traveled in west / time taken = 120/0.5 = 240 km/h.
so airspeed due west is = net speed - speed of plane = 240-220= 20 km/h.
airspeed due south is = distance traveled in west / time taken= 20/0.5= 40 km/h.
the magnitude of the wind velocity = √[(airspeed due south )² + (airspeed due west)²] = √ ( 40^2 + 20^2 ) = 44.72 km/h
the angle of airspeed south of west is tan⁻¹ ( airspeed due south / airspeed due west )= tan⁻¹(40/20)=63.43 degrees.
if wind velocity is 40 km/h due south, her velocity should have 20 km/h component in north.
so component west = sqrt ( 220^2 - 40^2 ) = 216.33 km/h.
the angle north of west is arctan( 40/216.33 ) = 10.47 degrees.
