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
We have two solenoids: solenoid 2 has twice the diameter, half the length, and twice as many turns as solenoid 1. The current in solenoid 2 is three times that in solenoid 1. How does the field B2 at the center of solenoid 2 compare to B1 at the center of solenoid 1?
Answer:
the field at the center of solenoid 2 is 12 times the field at the center of solenoid 1.
Explanation:
Recall that the field inside a solenoid of length L, N turns, and a circulating current I, is given by the formula:
[tex]B=\mu_0\, \frac{N}{L} I[/tex]
Then, if we assign the subindex "1" to the quantities that define the magnetic field ([tex]B_1[/tex]) inside solenoid 1, we have:
[tex]B_1=\mu_0\, \frac{N_1}{L_1} I_1[/tex]
notice that there is no dependence on the diameter of the solenoid for this formula.
Now, if we write a similar formula for solenoid 2, given that it has :
1) half the length of solenoid 1 . Then [tex]L_2=L_1/2[/tex]
2) twice as many turns as solenoid 1. Then [tex]N_2=2\,N_1[/tex]
3) three times the current of solenoid 1. Then [tex]I_2=3\,I_1[/tex]
we obtain:
[tex]B_2=\mu_0\, \frac{N_2}{L_2} I_2\\B_2=\mu_0\, \frac{2\,N_1}{L_1/2} 3\,I_1\\B_2=\mu_0\, 12\,\frac{N_1}{L_1} I_1\\B_2=12\,B_1[/tex]
The magnetic field at the center of solenoid 2 (B2) will be twelve times larger than the magnetic field at the center of solenoid 1 (B1), due to having four times the number of turns per unit length and three times the current.
Explanation:The magnetic field inside a solenoid is given by the formula B = µnI, where B is the magnetic field, µ (mu) is the magnetic permeability of the medium, n is the number of turns per unit length, and I is the current through the solenoid. Given solenoid 2 has twice the diameter of solenoid 1, half the length, and twice as many turns, with the current being three times that of solenoid 1, several factors will influence the magnetic field in solenoid 2 (B2) compared to solenoid 1 (B1).
The number of turns per unit length for solenoid 2 is four times that of solenoid 1, since it has twice as many turns and half the length. Additionally, the current in solenoid 2 is three times that in solenoid 1. Therefore, B2 will be twelve times larger than B1, since the magnetic field inside a solenoid is directly proportional to both the number of turns per unit length of the solenoid and the current through it (B2 = 12 * B1).
Approximately 80% of the energy used by the body must be dissipated thermally. The mechanisms available to eliminate this energy are radiation, evaporation of sweat, evaporation from the lungs, conduction, and convection. In this question, we will focus on the evaporation of sweat alone, although all of these mechanisms are needed to survive. The latent heat of vaporization of sweat at body temperature (37 °C) is 2.42 x 10^6 J/kg and the specific heat of a body is approximately 3500 J/(kg*°C).
(A) To cool the body of a jogger of mass 90 kg by 1.8°C , how much sweat has to evaporate?
O 130 g
O 230 g
O 23 g
O 13 g
Answer:
the correct answer is c) 23 g
Explanation:
The heat lost by the runner has two parts: the heat absorbed by sweat in evaporation and the heat given off by the body
Q_lost = - Q_absorbed
The latent heat is
Q_absorbed = m L
The heat given by the body
Q_lost = M [tex]c_{e}[/tex] ΔT
where m is the mass of sweat and M is the mass of the body
m L = M c_{e} ΔT
m = M c_{e} ΔT / L
let's replace
m = 90 3.500 1.8 / 2.42 10⁶
m = 0.2343 kg
reduced to grams
m = 0.2342 kg (1000g / 1kg)
m = 23.42 g
the correct answer is c) 23 g
230 g should have to evaporate.
Given that,
The latent heat of vaporization of sweat at body temperature (37 °C) is 2.42 x 10^6 J/kg and the specific heat of a body is approximately 3500 J/(kg*°C).The calculation is as follows:
[tex]= \frac{(90kg)(3500J/kg^{\circ})(1.8^{\circ}C)}{2.42\times 10^6J/kg}[/tex]
= 0.23 kg
= 230g
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The primary coil of a transformer has N1 = 275 turns, and its secondary coil has N2 = 2,200 turns. If the input voltage across the primary coil is Δv = (160 V)sin ωt, what rms voltage is developed across the secondary coil?
Answer:
Secondary voltage of transformer is 905.23 volt
Explanation:
It is given number of turns in primary of transformer [tex]N_1=275[/tex]
Number of turns in secondary [tex]N_2=2200[/tex]
Input voltage equation of the transformer
[tex]\Delta v=160sin\omega t[/tex]
Here [tex]v_{max}=160volt[/tex]
[tex]v_{rms}=\frac{160}{\sqrt{2}}=113.15volt[/tex]
For transformer we know that
[tex]\frac{V_1}{V_2}=\frac{N_1}{N_2}[/tex]
[tex]\frac{113.15}{V_2}=\frac{275}{2200}[/tex]
[tex]V_2=905.23Volt[/tex]
Therefore secondary voltage of transformer is 905.23 volt