Answer:
2 m = E / c^2 where m is mass of electron
E = h v where v is the frequency ( nu) of the incident photon
E = h c / y where y is the incident wavelength (lambda)
2 m = h / (c y)
y = h / (2 m c) wavelength required
y = 6.62 * 10E-34 / (2 * 9.1 * 10E-31 * 3 * 10E8) m
y = 3.31 / 27.3 E-11 m
y = 1.21 E -12 m = .0121 Angstrom units
Answer:
The neutral wire is often confused with ground wire, but in reality, they serve two distinct purposes. Neutral wires carry currents back to power source to better control and regulate voltage. Its overall purpose is to serve as a path to return energy.
Answer:
option C
Explanation:
given,
mass of the three planet is same
radius of the planets are
R₁ > R₂ > R₃
expression of escape velocity

G is the gravitational constant
M is the mass of the planet
R is the radius of the planet
from the above expression we can clearly conclude that the escape velocity is inversely proportional to the radius of the Planet.
radius of planet increases escape velocity decreases.
Hence planet 3 has the smallest radius so the escape velocity of the third planet will be maximum.
The correct answer is option C
Answer:
(a) 1.21 m/s
(b) 2303.33 J, 152.27 J
Explanation:
m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s
(a) Let their velocity after striking is v.
By use of conservation of momentum
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v
- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v
v = ( - 356.25 + 607.94) / 208 = 1.21 m /s
(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2
= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)
= 0.5 (1335.94 + 3270.7) = 2303.33 J
Kinetic energy after collision = 1/2 (m1 + m2) v^2
= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J
Answer:
The frictional torque is 
Explanation:
From the question we are told that
The mass attached to one end the string is 
The mass attached to the other end of the string is 
The radius of the disk is 
At equilibrium the tension on the string due to the first mass is mathematically represented as

substituting values


At equilibrium the tension on the string due to the mass is mathematically represented as



The frictional torque that must be exerted is mathematically represented as

substituting values

