Answer:
(a). The kinetic energy stored in the fly wheel is 46.88 MJ.
(b). The time is 1.163 hours.
Explanation:
Given that,
Radius = 1.50 m
Mass = 475 kg
Power 
Rotational speed = 4000 rev/min
We need to calculate the moment of inertia
Using formula of moment of inertia
Put the value into the formula
(a). We need to calculate the kinetic energy stored in the fly wheel
Using formula of K.E
Put the value into the formula
(b). We need to calculate the length of time the car could run before the flywheel would have to be brought backup to speed
Using formula of time
Hence, (a). The kinetic energy stored in the fly wheel is 46.88 MJ.
(b). The time is 1.163 hours.
Answer:
The final angular velocity is 20rad/s
Explanation:
We are given;
mass, m = 12 kg
radius, r = 0.25 m
Work done;W = 75 J
Moment of inertia of cylinder, I = (1/2) mr²
Thus,
I = (1/2) x 12 x 0.25² = 0.375 kg.m²
Now, from work energy theorem,
Work done = Change in kinetic energy
So, W = KE_f - KE_i
Now, Initial Kinetic Energy (KE_i) = 0
Final Kinetic Energy; KE_f = (1/2)Iω²
So, KE_f = (1/2) x 0.375 x ω²
KE_f = 0.1875 ω²
Now, W = 75 J
Thus,
From, W = KE_f - KE_i, we have;
75 = 0.1875 ω² - 0
75 = 0.1875 ω²
ω² = 75/0.1875
ω² = 400
ω = √400
ω = 20 rad/s
Hydrogen has one electron in its outermost shell, while fluorine has seven electron in its outermost shell, hence both hydrogen and fluorine needs a single electron to complete its outermost shell.
That's why there is a single bond between hydrogen and fluorine.
Hence both hydrogen and fluorine share one electron with each other, so option "A" is correct.
The magnetizing current in a transformer is rich in 3rd harmonic. This is because harmonics are AC voltages and currents with frequencies that are generally higher.
Answer:
2.083 V.
Explanation:
Stopping potential is the potential that is required to stop the current to zero . This potential is applied externally to oppose the potential created by the photoelectric effect . It gives the measure the photoelectric potential being generated .
Here current drops to 25 μA to 19 μA by a potential of 500mV
Change in current
= 25 - 19 = 6 μA
Voltage requirement for unit reduction in current
= 500 / 6 μA
To reduce current 0f 25 μA
requirement of V = (500 / 6 ) x 25 = 2083.33 mV = 2.083 V.
