Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Power = 1800W (or 1.8KW by dividing by 1000)
Time = 3 hours
Power = energy/ time
1.8KW = energy/ 3
x3
5.4Kw/h= energy
(5.4KJ or 5400J used)
$0.15 Kw/h
$0.15 X 5.4 = 0.81
Thus, cost $0.81
Hope this helps!
Explanation:
Let 
is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.
The magnetic force is balanced by the centripetal force acting on the proton as :
r is the radius of path,
Time period is given by :
Frequency of proton is given by :
The wavelength of radiation is given by :
So, the wavelength of radiation produced by a proton is 
. Hence, this is the required solution.
The total work is
(mass of the elevator, kg) x (9.8 m/s²) x (9.0 m) Joules .
Answer:
5.49×10⁻⁴ lbm
Explanation:
Convert volume to m³.
V = (200 cm³) (1 m / 100 cm)³ = 0.0002 m³
Find mass in kg.
m = ρV
m = (1.24507 kg/m³) (0.0002 m³)
m = 0.000249 kg
Convert mass to lbm.
m = (0.000249 kg) (2.205 lbm/kg)
m = 0.000549 lbm
m = 5.49×10⁻⁴ lbm
