A is not the correct answer because the amplitude and oscillation and there is 1/2 A in oscillation
Answer: did you get the answers?
Explanation:
Answer:
W= -2.5 (p₁*0.0012) joules
Explanation:
Given that p₀= initial pressure, p₁=final pressure, Vi= initial volume=0 and Vf=final volume= 6/5 liters where p₁=p₀ then
In adiabatic compression, work done by mixture during compression is
W=
where f= final volume and i =initial volume, p=pressure
p can be written as p=K/V^γ where K=p₀Vi^γ =p₁Vf^γ
W= 
W= K/1-γ ( 1/Vf^γ-1 - 1/Vi^γ-1)
W=1/1-γ (p₁Vf-p₀Vi)
W= 1/1-1.40 (p₁*6/5 -p₀*0)
W= -2.5 (p₁*6/5*0.001) changing liters to m³
W= -2.5 (p₁*0.0012) joules
Answer:
49.85 V
Explanation:
u = 0, s = 5.62 cm, t = 1.15 x 10^-6 s
Let the electric field is E and voltage is V.
Use second equation of motion
s = ut + 1/2 a t^2
5.62 x 10^-2 = 0 + 0.5 a x (1.15 x 10^-6)^2
a = 8.5 x 10^10 m/s^2
m x a = q x E
E = m x a / q
E = (1.67 x 10^-27 x 8.5 x 10^10) / (1.6 x 10^-19)
E = 887.19 V/m
V = E x s
V = 887.19 x 5.62 x 10^-2 = 49.85 V