The photoelectric emission is possible if the wavelength of the incident light is less than that of yellow light
Answer:
5.43 x 10^-3 Nm
Explanation:
N = 52.5, radius, r = 5.35 cm = 0.0535 m, B = 0.455 T, I = 25.3 mA = 0.0253 A
Torque = N I A B Sin theta
Here, theta = 90 degree
Torque = 52.5 x 0.0253 x 3.14 x 0.0535 x 0.0535 x 0.455
Torque = 5.43 x 10^-3 Nm
Answer:
a) Pb= 200 PA
b).work done= -3600 joules
c).3600joules
D).the system works under isothermal condition so no heat was transferred
Explanation:
2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.
a). PbVb= PaVa
Pb= (PaVa)/VB
Pb= (600*3)/9
Pb= 1800/9
Pb= 200 PA
b). work done= n(Pb-Pa)(Vb-Va)
Work done= 2*(200-600)(9-3)
Work done= -600(6)
Work done=- 3600 Pam³
work done= -3600 joules
C). Change in internal energy I the work done on the system
= 3600joules
D).the system works under isothermal condition so no heat was transferred
We did this experiment before, when the rope moves, it represents the waves passing through in from the level of intensity. I hope this is a good answer.
Answer:
m=417.24 kg
Explanation:
Given Data
Initial mass of rocket M = 3600 Kg
Initial velocity of rocket vi = 2900 m/s
velocity of gas vg = 4300 m/s
Θ = 11° angle in degrees
To find
m = mass of gas
Solution
Let m = mass of gas
first to find Initial speed with angle given
So
Vi=vi×tanΘ...............tan angle
Vi= 2900m/s × tan (11°)
Vi=563.7 m/s
Now to find mass
m = (M ×vi ×tanΘ)/( vg + vi tanΘ)
put the values as we have already solve vi ×tanΘ
m = (3600 kg ×563.7m/s)/(4300 m/s + 563.7 m/s)
m=417.24 kg
