The emf induced in the second coil is given by:
V = -M(di/dt)
V = emf, M = mutual indutance, di/dt = change of current in the first coil over time
The current in the first coil is given by:
i = i₀
i₀ = 5.0A, a = 2.0×10³s⁻¹
i = 5.0e^(-2.0×10³t)
Calculate di/dt by differentiating i with respect to t.
di/dt = -1.0×10⁴e^(-2.0×10³t)
Calculate a general formula for V. Givens:
M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)
Plug in and solve for V:
V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))
V = 320e^(-2.0×10³t)
We want to find the induced emf right after the current starts to decay. Plug in t = 0s:
V = 320e^(-2.0×10³(0))
V = 320e^0
V = 320 volts
We want to find the induced emf at t = 1.0×10⁻³s:
V = 320e^(-2.0×10³(1.0×10⁻³))
V = 43 volts
Answer:
Gold is more dense.
Gold sinks faster than lead. That's why gold is found in the bottom of a gold pan or river.
Answer:
the faster an object moves the more kinetic it has. the more mass an object has, the more kinetic energy it has.
Have wavelengths that are longer than normal.
Answer:
h=17357.9m
Explanation:
The atmospheric pressure is just related to the weight of an arbitrary column of gas in the atmosphere above a given area. So, if you are higher in the atmosphere less gass will be over you, which means you are bearing less gas and the pressure is less.
To calculate this, you need to use the barometric formula:
Where R is the gas constant, M the molar mass of the gas, g the acceleration of gravity, T the temperature and h the height.
Furthermore, the specific gas constant is defined by:
Therefore yo can write the barometric formula as:
at the surface of the planet (h =0) the pressure is ![P_0[\tex]. The pressure at the height requested is half of that:[tex]P=\frac{P_0}{2}](https://tex.z-dn.net/?f=P_0%5B%5Ctex%5D.%20The%20pressure%20at%20the%20height%20requested%20is%20half%20of%20that%3A%3C%2Fp%3E%3Cp%3E%5Btex%5DP%3D%5Cfrac%7BP_0%7D%7B2%7D)
applying to the previuos equation:
solving for h:
h=17357.9m
