The formula for the energy in a capacitor , u in terms of q and c is q²/2c
<h3>What is the energy of a capacitor?</h3>
The energy of a capacitor u = 1/2qv where
- q = charge on capacitor and
- v = voltage across capacitor.
<h3>What is the capacitance of a capacitor?</h3>
Also, the capacitance of a capacitor c = q/v where
- q = charge on capacitor and
- v = voltage across capacitor.
So, v = q/c
<h3>
The formula for energy of the capacitor in terms of q and c</h3>
Substituting v into u, we have
u = 1/2qv
= 1/2q(q/c)
= q²/2c
So, the formula for the energy in a capacitor , u in terms of q and c is q²/2c
Learn more about energy in a capacitor here:
brainly.com/question/10705986
#SPJ12
Answer:
15 m/s or 1500 cm/s
Explanation:
Given that
Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s
Distance moved during the hook, d(h) = 5 cm = 0.05 m
Distance moved by the fist, d(f) = 100 cm = 1 m
Average speed of the fist during the hook, v(f) = ? cm/s = m/s
This can be solved by a very simple relation.
d(f) / d(h) = v(f) / v(h)
v(f) = [d(f) * v(h)] / d(h)
v(f) = (1 * 0.75) / 0.05
v(f) = 0.75 / 0.05
v(f) = 15 m/s
Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s
Answer:
Acceleration of the second particle at that moment is given as
Explanation:
As we know that both cars are connected by same spring
So on this system of two cars there is no external force
So we will have
now we have
now we have
so we have
Kinetic energy = (1/2) (mass) (speed)²
BUT . . . in order to use this equation just the way it's written,
the speed has to be in meters per second. So we'll have to
make that conversion.
KE = (1/2) · (1,451 kg) · (48 km/hr)² · (1000 m/km)² · (1 hr/3,600 sec)²
= (725.5) · (48 · 1000 · 1 / 3,600)² (kg) · (km·m·hr / hr·km·sec)²
= (725.5) · ( 40/3 )² · ( kg·m² / sec²)
= 128,978 joules (rounded)
