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Answer:
Newtows first law of motion
Given:
m = 555 g, the mass of water in the calorimeter
ΔT = 39.5 - 20.5 = 19 °C, temperature change
c = 4.18 J/(°C-g), specific heat of water
Assume that all generated heat goes into heating the water.
Then the energy released is
Q = mcΔT
= (555 g)*(4.18 J/(°C-g)*(19 °C)
= 44,078.1 J
= 44,100 J (approximately)
Answer: 44,100 J
Answer:
Part a)

Part b)

Part c)

Part d)
Net force on a closed loop in uniform magnetic field is always ZERO

Explanation:
As we know that force on a current carrying wire is given as

now we have
Part a)
current in side 166 cm and magnetic field is parallel
so we have

here we know that L and B is parallel to each other so

Part b)
For 68.1 cm length wire we have

here we know that


so we have


Part c)
For 151 cm length wire we have

here we know that


so we have


Part d)
Net force on a closed loop in uniform magnetic field is always ZERO

Answer:
z = 0.8 (approx)
Explanation:
given,
Amplitude of 1 GHz incident wave in air = 20 V/m
Water has,
μr = 1
at 1 GHz, r = 80 and σ = 1 S/m.
depth of water when amplitude is down to 1 μV/m
Intrinsic impedance of air = 120 π Ω
Intrinsic impedance of water = 
Using equation to solve the problem

E(z) is the amplitude under water at z depth
E_o is the amplitude of wave on the surface of water
z is the depth under water



now ,


taking ln both side
21.07 x z = 16.81
z = 0.797
z = 0.8 (approx)