Answer:
Kinetic energy is given by:
K.E. = 0.5 m v²
Susan has mass, m = 25 kg
Velocity with which Susan moves is, v = 10 m/s
Hannah has mass, m' = 30 kg
Velocity with which Hannah moves is, v' = 8.5 m/s
<u>Kinetic energy of Susan:</u>
0.5 m v² = 0.5 × 25 kg × (10 m/s)² = 1250 J
<u>Kinetic energy of Hannah:</u>
0.5 m v'² = 0.5 × 30 kg × (8.5 m/s)² = 1083.75 J
Susan's kinetic energy is <u>1250 J </u>and Hannah's kinetic energy is <u>1083.75 J</u>.
Since kinetic energy is dependent on mass and square of speed. Thus, speed has a greater effect than mass. As it is evident from the above example. Susan has greater kinetic energy due to higher speed than Hannah.
The force of attraction between the two particles will remain the same, because when mass is doubled, force of attraction is doubled. However, when distance between their centers is doubled, then force of attraction is halved. As such double and half cancel out each other and force of attraction remains the same.
Answer with Explanation:
a.Intensity of radiation is directly proportional to the frequency of radiation
When the intensity of radiation increases then the frequency of radiation increases and therefore, the number of photo-electrons emitted by the metal increases.
b.When all of the wavelength in the radiation are increased by the same amount
We know that

Frequency is inversely proportional to the wavelength.
Therefore, the frequency decrease .
When the frequency decreases then the number of photo-electrons emitted by the metal decrease.
c.When the work function of the metal is increased then the gain of kinetic energy decreases .
When energy decreases then the number of photo-electrons emitted by the metal decreases.
Answer:
E.true only when no charge is enclosed within the Gaussian surface.
Explanation:
Because Gauss’s law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge.
Answer:
Explanation:
Given that, .
R = 12 ohms
C = 500μf.
Time t =? When the charge reaches 99.99% of maximum
The charge on a RC circuit is given as
A discharging circuit
Q = Qo•exp(-t/RC)
Where RC is the time constant
τ = RC = 12 × 500 ×10^-6
τ = 0.006 sec
The maximum charge is Qo,
Therefore Q = 99.99% of Qo
Then, Q = 99.99/100 × Qo
Q = 0.9999Qo
So, substituting this into the equation above
Q = Qo•exp(-t/RC)
0.9999Qo = Qo•exp(-t / 0.006)
Divide both side by Qo
0.9999 = exp(-t / 0.006)
Take In of both sodes
In(0.9999) = In(exp(-t / 0.006))
-1 × 10^-4 = -t / 0.006
t = -1 × 10^-4 × - 0.006
t = 6 × 10^-7 second
So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge