# Questions On Physics

Submitted By anmolshienh
Words: 1083
Pages: 5

55.a) v of ball= 300-20= 280 m/s Speed of light= 3*10^8 m/s b) I would not expect my watch to be affected by the time dilation because the plane does not move fast enough. The watch would not be able to pick up on a small change in time dilation. 56.

No, because the time dilation A will be ahead of B.
b) Δt’= Δt/√(1-v^2/c^2) = Δt/√(1-100^2/7.0*10^8^2) = Δt/0.9999, very little difference
If c= 2v
Δt’= Δt/√(1-100^2/200^2) = Δt/√0.75 = Δt/ 0.966; a greater difference of about 13%
Therefore if the light was twice the speed of the satellite it would increase the magnitude by which time is dilated and Δt’ would be a larger number resulting in “slower time”.

57. a) lifetime= 2.20*10^-6 v= 0.99c Δ t= Δt˳ / ( 1-v^2/c^2)^1/2
Δ t= 2.2*10^-6 / ( 1-0.99^2)^1/2
Δt = 1.6*10^-5

D= vΔt= 0.99(3*10^8 m/s) ( 1.6*10^-6 s)
D= 4752 m
Therefore, it travels 4752 m before decaying.
b) Δ t earth = 4752/ 0.99*3*10^8 = 1.6*10^-5 s = 16.0 µs = 1.6*10^-5 s
c) Δ t muons= Δ t earth/ γ = 16/7.09 = 2.25 µs dM= v (Δ tp) = v Δ t earth/ γ = 0.99*16/7.09 =2.23 m
d) This is supportive evidence for the theory of relativity as it shows that when an object reaches the speed of light time dilation is effective.
58) Length of spaceship A = 60.0 m Length of spaceship B= 120.0 m Mass of A = 15000 kg Speed of spaceship A= 0.70 c
a) Lm= Ls √(1-v^2/c^2)
Lm= 60 (√(1-(0.70c)^2/c^2)
Lm=42.8 m
Therefore the rocket is 42.8 m long.
b) 1/ v (1-(0.7)^2) = 1.4
1.4*2= 2.8 v =√1-1/2.8^2 *c = 0.93c
Therefore the speed is 0.93 c.

c) Mm=ms/√(1-v^2/c^2) = 15000 kg/ √(1-(0.70c)^2/c^2) = 21004.2 kg
59) 30 days* 24h/day* 3600s/h=2592000s
ΔE= p Δt
ΔE= (2.00*10^9 W) (2592000s)
ΔE = 7.77*10^15 J

ΔE = Δmc^2
7.77*10^15= Δm (3.00*10^8)^2
Δm= 8.63*10^-2 kg
Therefore, 8.63*10^-2 kg is how much mass the plant converts into energy in one month.
60) a) E total= mc^2 /√(1-v^2/c^2)
E total= (1.67*10^-27 kg) (3.00*10^8 m/s)^2/√(1- 0.99980001)
E total= 1.07*10^-8 J

E total= E rest+ Ek
Ek= E total – E rest
Ek= mc^2 /√(1-v^2/c^2)- mc^2
Ek= 1.07*10^-8 J – (1.67*10^-27 kg) (3.00*10^8 m/s)^2
Ek= 1.05 *10^-8 J

b) E total= E rest + Ek 1.07*10^-8 J = E rest + 1.05*10^-8 J
E rest = 1.07*10^-8 J - 1.05*10^-8 J
E rest = 2.0 *10^-10 J
c) No, the particle accelerator will never be able to accelerate any particle to the speed of light because the closer it gets to that speed it’ll need more energy to keep moving up to it; kinetic energy is greater hence, it’s difficult to accelerate.
61. a) E= hc/ λ E= (6.63*10^-34 Js) (3.00 * 10^8 m/s) / 5.90* 10^-7 m E= 3.37* 10^-19 J
Therefore, the energy of a quantum of light with a wavelength of 590 nm is 3.37*10^-19 J. b) 3.0 eV* 1.6*10^-19= 6.0*10^-19
Therefore, the color of light will not be able to produce photoelectrons from the surface of the metal as the work function (6.0*10^-19 J) is greater then the energy obtained (3.37*10^-19 J).
62) W= 1.5eV * 1.60*10^-19 J/eV W= 2.3*10^-19 J 630/10^9= λ λv= c 6.3*10^-7 * v= 3.00*10^8 v = 4.76*10^-14 Hz E k= hf-w E k= (6.63*10^-34 Js) (4.76*10^14 Hz) – 2.4*10^-19J) E k = 7.5*10^-18 J Therefore, the