Answer:
19.1 deg
Explanation:
v = speed of the proton = 8 x 10⁶ m/s
B = magnitude of the magnetic field = 1.72 T
q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C
F = magnitude of magnetic force on the proton = 7.20 x 10⁻¹³ N
θ = Angle between proton's velocity and magnetic field
magnitude of magnetic force on the proton is given as
F = q v B Sinθ
7.20 x 10⁻¹³ = (1.6 x 10⁻¹⁹) (8 x 10⁶) (1.72) Sinθ
Sinθ = 0.327
θ = 19.1 deg
Resistance is what Inertia goes to.....
Answer:
A + B = C Ax = 2 Ay = 0 Bx = 0 By = 6
Ax + Bx = Cx = 2
Ay + By = Cy = 6
C = (2^2 + 6^2)^1/2 = 6.32
Tan Cy / Cx = 6 / 2 = 3
Cy at 71.6 deg
Answer:
Explanation:
(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)
ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .
ΔK = 0
ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .
ΔUg = positive
ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .
ΔUs = 0
ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence
ΔEth = negative .
b )
work done by athlete
= 400 x 2 = 800 J
energy output = 800 J
c )
It is 25% of metabolic energy output of his body
so metalic energy output of body
= 4x 800 J .
3200 J
power = energy output / time
= 3200 / 1.6
= 2000 W .
d )
1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .
2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .
