The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval,
:
First we need to find the net charge flowing at a certain point of the wire in one second,
. Using I=0.92 A and re-arranging the previous equation, we find
Now we know that each electron carries a charge of
, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
<span>No, because the truck applies more pressure than the bridge can support.</span>
The answer to the question is A
According to the Law of Conservation of Energy, energy is neither created nor destroyed. They are just transferred from one system to another. To obey this law, the energy of the substances inside the container must be equal to the substance added to it. The energy is in the form of heat. There can be two types of heat energy: latent heat and sensible heat. Sensible heat is energy added or removed when a substance changes in temperature. Latent heat is the energy added or removed at a constant temperature during a phase change. Since there is no mention of phase change, we assume the heat involved here is sensible heat. The equation for sensible heat is:
H = mCpΔT
where
m is the mass of the substance
Cp is the specific heat of a certain type of material or substance
ΔT is the change in temperature.
So the law of conservation of heat tells that:
Sensible heat of Z + Sensible heat of container = Sensible heat of X
Since we have no idea what these substances are, there is no way of knowing the Cp. We can't proceed with the calculations. So, we can only assume that in the duration of 15 minutes, the whole system achieves equilibrium. Therefore, the equilibrium temperature of the system is equal to 32°C. The answer is C.
Answer:
θ = 36.2º
Explanation:
When light passes through a polarizer it becomes polarized and if it then passes through a second polarizer, it must comply with Malus's law
I = I₀ cos² tea
The non-polarized light between the first polarized of this leaves half the intensity, with vertical polarization
I₁ = I₀ / 2
I₁ = 845/2
I₁ = 422.5 W / m²
In this case, the incident light in the second polarizer has an intensity of I₁ = 422.5 W / m² and the light that passes through the polarizer has a value of
I = 275 W / m
²
Cos² θ = I / I₁
Cos θ = √ I / I₁
Cos θ = √ (275 / 422.5)
Cos θ = 0.80678
θ = cos⁻¹ 0.80678
θ = 36.2º
This is the angle between the two polarizers