The frequency of a wave is the number of waves that passes through a point in a certain time. The less waves that pass in a period of time the lower the frequency of the wave. The more waves that pass in a period of time the higher the frequency of the wave. When measuring wave length the time period used is usually one second.
Answer:
(a) 17.37 rad/s^2
(b) 12479
Explanation:
t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0
w = v / r = 99 / 0.06 = 1650 rad/s
(a) Use first equation of motion for rotational motion
w = w0 + α t
1650 = 0 + α x 95
α = 17.37 rad/s^2
(b) Let θ be the angular displacement
Use third equation of motion for rotational motion
w^2 = w0^2 + 2 α θ
1650^2 = 0 + 2 x 17.37 x θ
θ = 78367.87 rad
number of revolutions, n = θ / 2 π
n = 78367.87 / ( 2 x 3.14)
n = 12478.9 ≈ 12479
Answer:
diameter of largest orbit is 0.60 m
Explanation:
given data
isotopes accelerates KE = 6.5 MeV
magnetic field B = 1.2 T
to find out
diameter
solution
first we find velocity from kinetic energy equation
KE = 1/2 × m×v² ........1
6.5 × 1.6 × 
= 1/2 × 1.672 × 
×v²
v = 3.5 × 
m/s
so
radius will be
radius = 
........2
radius = 
radius = 0.30
so diameter = 2 × 0.30
so diameter of largest orbit is 0.60 m
Answer: Escaped volume = 0.0612m^3
Explanation:
According to Boyle's law
P1V1 = P2V2
P1 = initial pressure in the tire = 36.0psi + 14.696psi = 50.696psi (guage + atmospheric pressure)
P2 = atmospheric pressure= 14.696psi
V1 = volume of tire =0.025m^3
V2 = escaped volume + V1 ( since air still remain in the tire)
V2 = P1V1/P2
V2 = 50.696×0.025/14.696
V2 = 0.0862m^3
Escaped volume = 0.0862 - 0.025 = 0.0612m^3
