Answer:
doubled the initial value
Explanation:
Let the area of plates be A and the separation between them is d.
Let V be the potential difference of the battery.
The energy stored in the capacitor is given by
U = Q^2/2C ...(1)
Now the battery is disconnected, it means the charge is constant.
the separation between the plates is doubled.
The capacitance of the parallel plate capacitor is inversely proportional to the distance between the plates.
C' = C/2
the new energy stored
U' = Q^2 / 2C'
U' = Q^2/C = 2 U
The energy stored in the capacitor is doubled the initial amount.
Solution :
Let
kg
m/s
Let
and
are the speeds of the disk
and
after the collision.
So applying conservation of momentum in the y-direction,
![$0=m_1 .v_1_y -m_2 .v_2_y $](https://tex.z-dn.net/?f=%240%3Dm_1%20.v_1_y%20-m_2%20.v_2_y%20%24)
![$v_1_y = v_2_y$](https://tex.z-dn.net/?f=%24v_1_y%20%3D%20v_2_y%24)
![$v_1 . \sin 60=v_2. \sin 30$](https://tex.z-dn.net/?f=%24v_1%20.%20%5Csin%2060%3Dv_2.%20%5Csin%2030%24)
![$v_2 = v_1 \times \frac{\sin 60}{\sin 30}$](https://tex.z-dn.net/?f=%24v_2%20%3D%20v_1%20%5Ctimes%20%5Cfrac%7B%5Csin%2060%7D%7B%5Csin%2030%7D%24)
![$v_2=1.732 \times v_1$](https://tex.z-dn.net/?f=%24v_2%3D1.732%20%5Ctimes%20v_1%24)
Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.
Now applying conservation of momentum in the x-direction,
![$m_1.u_1=m_1.v_1_x+m_2.v_2_x$](https://tex.z-dn.net/?f=%24m_1.u_1%3Dm_1.v_1_x%2Bm_2.v_2_x%24)
![$u_1=v_1_x+v_2_x$](https://tex.z-dn.net/?f=%24u_1%3Dv_1_x%2Bv_2_x%24)
![$5=v_1. \cos 60 + v_2 . \cos 30$](https://tex.z-dn.net/?f=%245%3Dv_1.%20%5Ccos%2060%20%2B%20v_2%20.%20%5Ccos%2030%24)
![$5=v_1. \cos 60 + 1.732 \times v_1 \cos 30$](https://tex.z-dn.net/?f=%245%3Dv_1.%20%5Ccos%2060%20%2B%201.732%20%5Ctimes%20v_1%20%5Ccos%2030%24)
m/s
So, ![$v_2 = 1.732 \times 2.5$](https://tex.z-dn.net/?f=%24v_2%20%3D%201.732%20%5Ctimes%202.5%24)
= 4.33 m/s
Therefore, speed of the disk 2 after collision is 4.33 m/s
The acceleration of gravity on Earth is 9.8 m/s² .
The speed of a falling object keeps increasing smoothly,
in such a way that the speed is always 9.8 m/s faster than
it was one second earlier.
If you 'drop' the penny, then it starts out with zero speed.
If you also start the clock at the same instant, then
After 1.10 sec, Speed = (1.10 x 9.8) = 10.78 meters/sec
After 1.85 sec, Speed = (1.85 x 9.8) = 18.13 meters/sec
But you want this second one given in a different unit of speed.
OK then:
= (18.13 meter/sec) x (3,600 sec/hr) x (1 mile/1609.344 meter)
= (18.13 x 3,600 / 1609.344) (mile/hr) = 40.56 mph (rounded)
We did notice that in an apparent effort to make the question
sound more erudite and sophisticated, you decided to phrase
it in terms of 'velocity'. We can answer it in those terms, if we
ASSUME that there is no wind, and the penny therefore doesn't
acquire any horizontal component of motion on its way down.
With that assumption in force, we are able to state unequivocally
and without fear of contradiction that each 'speed' described above ...
with the word 'downward' appended to it ... does become a 'velocity'.