Answer:
-0.038 N
Explanation:
F=K Q1 Q2/r^2 by COULOMB'S LAW
F= 9×10^9×1×10^-5×-1.5×10^-5/(6)^2
F= -0.038 N
<span>Ik it has something to do with not always being able to be used. Example: Goes dark at night therefore no sunlight some people say a but i would say d but the person that said it was a was not very trustable so yea i would go with D hope this helped:)</span>
Apply Gay-Lussac's law:
P/T = const.
P = pressure, T = temperature, the quotient of P/T must stay constant.
Initial P and T values:
P = 180kPa, T = -8.0°C = 265.15K
Final P and T values:
P = 245kPa, T = ?
Set the initial and final P/T values equal to each other and solve for the final T:
180/265.15 = 245/T
T = 361K
Answer:
Explanation:
v = Speed of electron = 
(generally the order of magnitude is 6)
m = Mass of electron = 
Work done would be done by
The work required to stop the electron is
Answer:
Explanation:
Given a square side loop of length 10cm
L=10cm=0.1m
Then, Area=L²
Area=0.1²
Area=0.01m²
Given that, frequency=60Hz
And magnetic field B=0.8T
a. Flux Φ
Flux is given as
Φ=BA Sin(wt)
w=2πf
Φ=BA Sin(2πft)
Φ=0.8×0.01 Sin(2×π×60t)
Φ=0.008Sin(120πt) Weber
b. EMF in loop
Emf is given as
EMF= -N dΦ/dt
Where N is number of turns
Φ=0.008Sin(120πt)
dΦ/dt= 0.008×120Cos(120πt)
dΦ/dt= 0.96Cos(120πt)
Emf=-NdΦ/dt
Emf=-0.96NCos(120πt). Volts
c. Current induced for a resistance of 1ohms
From ohms law, V=iR
Therefore, Emf=iR
i=EMF/R
i=-0.96NCos(120πt) / 1
i=-0.96NCos(120πt) Ampere
d. Power delivered to the loop
Power is given as
P=IV
P=-0.96NCos(120πt)•-0.96NCos(120πt)
P=0.92N²Cos²(120πt) Watt
e. Torque
Torque is given as
τ=iL²B
τ=-0.96NCos(120πt)•0.1²×0.8
τ=-0.00768NCos(120πt) Nm
