1.cool down
2.activity log
3.specific warm up
4.activities of daily living
5.planned exercise
6.general warm up
Answer:
I= 3.5 amps
Explanation:
Step one:
given data
rating of resistor R= 8 ohms
power P= 100W
Required
The current I
Step two
Yet this power is also given by

make I subject of the formula we have

substitute

Answer:
pahingi po ng pic pls para masagutang kopo iyan
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