The formula for the energy stored in the magnetic field of an inductor is:
E = (1/2) (inductance) (current)² .
In the present situation:
Energy = (3 kilo-watt-hour) x (1,000 / kilo) x (joule/watt-sec) x (3,600 sec/hr)
= (3 · 1000 · 3,600) (kilo·watt·hr·joule·sec / kilo·watt·sec·hr)
= 1.08 x 10⁷ joules .
Now to find the inductance:
E = (1/2) (inductance) (current)²
(1.08 x 10⁷ joules) = (1/2) (inductance) (300 Amp)²
(2.16 x10⁷ joules) = (inductance) (300 Amp)²
Inductance = (2.16 x10⁷ joules) / (300 Amp)²
= 2.16 x10⁷ / 90,000 Henrys
I get 240 Henrys .
This is a big inductance. Possibly the size of your house.
To get a big inductance, you want to wind the coil
with a huge number of turns of very fine wire, in
a small space.
In this case, however, if you plan on running 300A through
your coil, it'll have to be wound with a very thick conductor ...
like maybe 1/4-inch solid copper wire, or even copper tubing,
You have competing requirements.
There are cheaper, easier, better ways to store 3 kWh of energy.
In fact, a quick back-of-the-napkin calculation says that
3 or 4 car batteries will do the job nicely.
Answer:
The height of the hill is, h = 38.42 m
Explanation:
Given,
The horizontal velocity of the soccer ball, Vx = 15 m/s
The range of the soccer ball, s = 42 m
The projectile projected from a height is given by the formula
S = Vx [Vy + √(Vy² + 2gh)] / g
Therefore,
h = S²g/2Vx² (Since Vy = 0)
Substituting the values
h = 42² x 9.8/ (2 x 15²)
= 38.42 m
Hence, the height of the hill is, h = 38.42 m
5.4 x 1014Hz
wavelength x frequency = the speed of light
Answer:
= 17º C
Explanation:
This is a calorimetry problem, where heat is yielded by liquid water, this heat is used first to melt all ice, let's look for the necessary heat (Q1)
Let's reduce the magnitudes to the SI system
Ice m = 80.0 g (1 kg / 1000 g) = 0.080 kg
L = 3.33 105 J / kg
Water M = 860 g = 0.860 kg
= 4186 J / kg ºC
Q₁ = m L
Q₁ = 0.080 3.33 10⁵
Q₁ = 2,664 10⁴ J
Now let's see what this liquid water temperature is when this heat is released
Q = M ΔT = M (T₀₁ -
)
Q₁ = Q
= T₀₁ - Q / M ce
= 26.0 - 2,664 10⁴ / (0.860 4186)
= 26.0 - 7.40
= 18.6 ° C
The initial temperature of water that has just melted is T₀₂ = 0ª
The initial temperature of the liquid water is T₀₁= 18.6
m + M 
= M T₀₁ - m T₀₂o2
= (M To1 - m To2) / (m + M)
= (0.860 18.6 - 0.080 0) / (0.080 + 0.860)
= 17º C
gg
