Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
Answer: Option (d) is correct.
Explanation:
Given, 1,152 British thermal units
1 British thermal unit = 1055.06 joules
So, in 1,152 British thermal units there will be :
Hence, from the given options the closest answer is of option (d). So, option (d) is correct.
<em>Resultant angle; θ = 25.59° </em>
This question is dealing with bearings and distance.
We are told that from point A, the camel walks 20 km at 15° in the south of east direction.
Thus, d_s,e = 20 km
Resolving along the horizontal east direction gives; d_e = 20 cos 15
d_e = 19.32 km
Also, resolving along the vertical south direction gives; d_s = 20 sin 15
d_s = 5.18 km
Net vertical distance; d_vert = 8km - 5.18km = 2.72 km
Net horizontal distance; d_hor = 25km - 19.32 km = 5.68 km
Now, the resultant angle is given by;
tan θ = d_vert/d_hor
tan θ = 2.72/5.68
tan θ = 0.4789
θ = tan^(-1) 0.4789
θ = 25.59°
Read more at; brainly.com/question/22518031
I beleive that the answer is B.
