Answer:
E =230.4 MJ
Explanation:
As 1 mole of electron = 6X 10^23 particles.
charge of an electron is 1.6 X 10 ^-19 C
Finding Charge:
(6X10^23 ) (2.7)(1.6X10^-19 C)
i.e. 192 K C
now to find the energy released from electrons
V=E/q
E=V X q
i.e E = 120 V X 192 K C
E =230.4 MJ
Answer:
The work required is -515,872.5 J
Explanation:
Work is defined in physics as the force that is applied to a body to move it from one point to another.
The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.
Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).
The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:


In this case:
- W=?
- m= 2,145 kg
- v2= 12

- v1= 25

Replacing:

W= -515,872.5 J
<u><em>The work required is -515,872.5 J</em></u>
Answer:
h = 9.57 seconds
Explanation:
It is given that,
Initial speed of Kalea, u = 13.7 m/s
At maximum height, v = 0
Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :




t = 1.39 s
Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

Here, a = -g


h = 9.57 meters
So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.
Answer: i) 2.356 × 10^-3 m = 2.356mm, ii) 4.712 × 10^-3 m = 4.712mm
Explanation: The formulae that relates the position of a fringe from the center to the wavelength, distance between slits and distance between slits and screen is given below as
y = R×(mλ/d)
Where y = distance between nth fringes and the center fringe.
m = order of fringe
λ = wavelength of light = 589nm = 589×10^-9m
R = distance between slits and screen = 1.0m
d = distance between slits = 0.25mm = 0.00025m
For distance between the first dark fringe and the center fringe.
This implies that m = 1
y = 1 × 589×10^-9 × 1/0.00025
y = 589×10^-9/0.00025
y = 2,356,000 × 10^-9
y = 2.356 × 10^-3 m = 2.356mm
For the second dark fringe, this implies that m = 2
y = 1 × 2 × 589×10^-9/0.00025
y = 1178 × 10^-9 /0.00025
y = 4,712,000 × 10^-9
y = 4.712 × 10^-3 m = 4.712mm