Consider taking a time-lapse video of an analog clock that is missing the hour hand. Assume that one frame of video is taken eve
1 answer:
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Answer:
a)11.25 J
b)Number of revolution = 1
Explanation:
Given that
Radius ,r= 0.8 m
m= 0.3 kg
Initial speed ,u= 10 m/s
final speed ,v= 5 m/s
a)
Initial energy
KEi= 15 J
Final kinetic energy
KEf=3.75 J
The energy transformed from mechanical to internal = 15 - 3.75 J = 11.25 J
b)
The minimum value to complete the circular arc
Now by putting the values
V= 2.82 m/s
So kinetic energy KE
KE=1.19 J
ΔKE= KEi - KE
ΔKE= 15- 1.19 J
ΔKE=13.80 J
The minimum energy required to complete 2 revolutions = 2 x 11.25 J
= 22.5 J
Here 22.5 J is greater than 13.8 J.So the particle will complete only one revolution.
Number of revolution = 1
Explanation:
Let us assume that the separation of plate be equal to d and the area of plates is . As the capacitance of capacitor is given as follows.
C =
It is known that the dielectric strength of air is as follows.
E =
Expression for maximum potential difference is that the capacitor can with stand is as follows.
dV = E × d
And, maximum charge that can be placed on the capacitor is as follows.
Q = CV
=
=
=
=
or, = 10.62 nC
Thus, we can conclude that charge on capacitor is 10.62 nC.
Answer:
E = 20.03 J
Explanation:
Given that,
The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C,
Voltage, V = 12 V
We need to find the energy delivered to the lightbulb filament during 2.00 s.
The energy delivered is given by :
. ....(1)
As,
As per Ohm's law, V = IR
Using formula (1).
So, the energy delivered to the lightbulb filament is 20.03 J.