Answer:
The work required to move this charge is 0.657 J
Explanation:
Given;
magnitude of charge, q = 4.4 x 10⁻⁶ C
Electric field strength, E = 3.9 x 10⁵ N/C
distance moved by the charge, d = 50 cm = 0.5m
angle of the path, θ = 40°
Work done is given as;
W = Fd
W = FdCosθ
where;
F is the force on the charge;
According the coulomb's law;
F = Eq
F = 3.9 x 10⁵ x 4.4 x 10⁻⁶ = 1.716 N
W = FdCosθ
W = 1.716 x 0.5 x Cos40
W = 0.657 J
Therefore, the work required to move this charge is 0.657 J
Answer:
Cost to leave this circuit connected for 24 hours is $ 3.12.
Explanation:
We know that,
f = frequency (60 Hz)
c= capacitor (10 µF = 
)
Substitute the given values
Given that, R = 200 Ω
X = 332.31 Ω
Current (I) = 0.361 amps
“Real power” is only consumed in the resistor,
In one hour 26 watt hours are used.
Energy used in 54 hours = 26 × 24 = 624 watt hours
E = 0.624 kilowatt hours
Cost = (5)(0.624) = 3.12
Applying diffraction equations;
d = 0.01/Number of lines; where 0.01 m = 1 cm, and d = spacing between lines
Therefore,
d = 0.01/2000 = 5*10^-6 m
Additionally,
d*Sin x = m*y; where x = Angle, m = order = 1, y = wavelength = 520 nm =520*10^-9 m
Substituting;
Sin x = my/d = (1*520*10^-9)/(5*10^-6) = 0.1040
x = Sin ^-1(0.104) = 5.97°
Therefore, first-order maximum for 520 nm will be 5.97°.
-- The potential energy of a 12-lb bowling ball up on the shelf
doesn't have anything to do with the temperature of the ball or
the shelf.
-- The potential energy of a jar full of gas does depend on the
temperature of the gas. The warmer it is, the greater its pressure
is, and the more work it can do if you let it out through a little hole
in the jar. If it gets hot enough, it'll have enough potential energy
to blow the jar to smithereens.
