Answer:0 J
Explanation:
Given
For first step
change in internal Energy of the system is 
Work done on the system 
For second step
change in internal Energy of the system is 
Work done on the system 
Work done on the system is considered as Positive and vice-versa.
and from first law of thermodynamics
for first step
overall heat added
For overall Process Heat added is 0 J
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
<span>If two objects collide, each object exerts a force equal to and in the opposite direction of the other.</span>
A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)
<h3>What is power?</h3>
In physics, power (P) is the work (W) done over a period of time.
- Step 1. Calculate the work done by the bodybuilder each time.
The bodybuilder lifts a 215 kg (m) weight to a height of 0.90 m (h). Being the gravity (g) of 9.81 m/s², we can calculate the work done in each lift using the following expression.
W = m × g × h = 215 kg × 9.81 m/s² × 0.90 m = 1.9 × 10³ N
- Step 2. Calculate the work done by the bodybuilder over 10 times.
W = 10 × 1.9 × 10³ N = 1.9 × 10⁴ N
- Step 3. Calculate the power exerted by the bodybuilder.
The bodybuilder does a work of 1.9 × 10⁴ N in a 45-s span.
P = 1.9 × 10⁴ N/45 s = 421 W
A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)
Learn more about power here: brainly.com/question/911620
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Answer:
Explanation:
Given:
Mass of the boat, m = 4040 kg
The driving force of engine, FB = 4660 N in west = + 4660 N
The force of wind, Fwi = 880 N in east = -880 N
The force of water, Fwa = 1400 N in east = -1400N
Total three forces are acting on the boat
Fnet= Fb+fwi+Fwa
Fnet= 4660 - 880 - 1400
Fnet= +2380N
Acceleration (a) = Force/mass
= 2380/4040
= 0.59m/s2
