Answer:
Number of turns per metre, n= 500/0.3= (5000/3)m^-1
Cross sectional areaof the square loop of wire, A= (0.1^2)m^2= 0.01m^2dB/dt= μn(dI/dt)= (4.00π x10^-7)(5000/3)(0.7)= 1.46608x10^-3T/s
The induced emf in the square loop of wire, ε= the rate of change of magnetic flux of the square loop of wire(dΦ/dt)= A(dB/dt)= (0.01)(1.46608x10^-3)= 0.0146608x10^-5VA
current flows in the square loop of wire since a potential difference(induced emf in this case) exists. Its magnitude,
I= ε/R where R is the resistance of the square loop of wire.
I= (0.0146608x10^-5)/30= 4.89x10^-7A
Answer:
2.06 m³/s
Explanation:
diameter of pipe, d = 0.81 m
diameter of constriction, d' = 0.486 m
radius, r = 0.405 m
r' = 0.243 m
density of oil, ρ = 821 kg/m³
Pressure in the pipe, P = 7970 N/m²
Pressure at the constriction, P' = 5977.5 N/m²
Let v and v' is the velocity of fluid in the pipe and at the constriction.
By use of the equation of continuity
A x v = A' x v'
r² x v = r'² x v'
0.405 x 0.405 x v = 0.243 x 0.243 x v'
v = 0.36 v' .... (1)
Use of Bernoulli's theorem

7970 + 0.5 x 821 x 0.36 x 0.36 x v'² = 5977.5 + 0.5 x 821 x v'² from (i)
1992.5 = 357.3 v'²
v' = 5.58 m/s
v = 0.36 x 5.58
v = 2 m/s
Rate of flow = A x v = 3.14 x 0.405 x 0.405 x 2 x 2 = 2.06 m³/s
Thus the rate of flow of volume is 2.06 m³/s.
Answer:
Hello there Dude answer is B :D hope it helped mark me brainliest.
<span>Answer:The weight of the door creates a CCW torque given by
Tccw = 145 N*3.13 m / 2
You need a CW torque that's equal to that
Tcw = F*2.5 m*sin20</span>