DE which is the differential equation represents the LRC series circuit where
L d²q/dt² + Rdq/dt +I/Cq = E(t) = 150V.
Initial condition is q(t) = 0 and i(0) =0.
To find the charge q(t) by using Laplace transformation by
Substituting known values for DE
L×d²q/dt² +20 ×dq/dt + 1/0.005× q = 150
d²q/dt² +20dq/dt + 200q =150
Work is done. work=forcexdisplacement. the ice skater glides 2 meters (displacement), so yes.
The magnitude of work done by the gas is 279 J and the sign is negative so W = -279 J as work is done by the system.
<u>Explanation:</u>
According to first law of thermodynamics, the change in internal energy of the system is equal to the sum of the heat energy added or released from the system with the work done on or by the system. If the heat energy is added to the system to perform a certain work, then the heat energy is taken as positive, while it will be negative when the heat energy is released from the system.
Similarly, in this case, the heat energy of 597 J is added to the system. So the heat energy will be positive, while the gas expansion occurs means work is done by the system.
ΔU = Q+W
Since ΔU is the change in internal energy which is given as 318 J and the heat energy added to the system is Q = 597 J.
Then the work done by the gas = ΔU - Q = 318 J - 597 J = - 279 J.
As the work is done by the system, so it will be denoted in negative sign and the magnitude of work done by the gas is 279 J.
To solve this problem we will resort to the concept of angle of incidence and refraction.
Since it is a reflection on a mirror, the angle provided for refraction will be equal to that of the incidence, that is, 25 °
The angle of reflation is always perpendicular to the surface so it is necessary to find the angle with respect to it.
Therefore the angle of the reflected beam of light made with the surface normal is 65°
Answer:
Explanation:
Let 100 m/s be the velocity of projection.
So horizontal component
= 100 cos42
= 74.31 m /s
Vertical component = - 100 sin 42 . in upward direction
66.91 m/s
Net displacement = 2.1 downwards ( + ve )
Using s = ut + 1/2 gt²
2.1 = - 66.91 t + .5 x 9.8 x t²
4.9 t² - 66.91 t - 2.1 = 0
t = 13.685 s
Horizontal distance covered
= 13.685 x 74.31
= 1016.93 m
If angle of projction is 40°
So horizontal component
= 100 cos40
= 76.60 m /s
Vertical component = - 100 sin 42 . in upward direction
64.27 m/s
Net displacement = 2.1 downwards ( + ve )
Using s = ut + 1/2 gt²
2.1 = -76.60 t + .5 x 9.8 x t²
4.9 t² - 76.60 t - 2.1 = 0
t = 15.659 s
Horizontal distance covered
= 15.659 x 76.60
= 1199.49 m
So horizontal range is increased , if angle of projection is increased .
