Answer:
The magnetic field exerts net force and net torque on the loop.
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
A barge floating in fresh water (p = 1000 kg/m3) is shaped like a hollow rectangular prism with base area A = 550 m2 and height H = 2.0 m. When empty the
bottom of the barge is located H0 = 0.45 m below the surface of the water. When fully
loaded with coal the bottom of the barge is located H; = 1.05 m below the surface. Part (a) Find the mass of the coal in kilograms.
Numeric : A numeric value is expected and not an expression.
m1 =
Part (b) How far would the barge be submerged (in meters) if m; = 450000 kg of coal had been placed on the empty barge? Numeric : A numeric value is expected and not an expression.
Answer:
a) [tex]\Delta m = 330000\,kg[/tex], b) [tex]h = 1.268\,m[/tex]
Explanation:
a) According the Archimedes' Principle, the buoyancy force is equal to the displaced weight of surrounding liquid. The mass of the coal in the barge is:
[tex]\Delta m \cdot g = \rho_{w}\cdot g \cdot \Delta V[/tex]
[tex]\Delta m = \rho_{w}\cdot \Delta V[/tex]
[tex]\Delta m = \left(1000\,\frac{kg}{m^{3}} \right)\cdot (550\,m^{2})\cdot (1.05\,m-0.45\,m)[/tex]
[tex]\Delta m = 330000\,kg[/tex]
b) The submersion height is found by using the equation derived previously:
[tex]\Delta m = \rho_{w}\cdot \Delta V[/tex]
[tex]450000\,kg = \left(1000\,\frac{kg}{m^{3}}\right)\cdot (550\,m^{2})\cdot (h-0.45\,m)[/tex]
The final submersion height is:
[tex]h = 1.268\,m[/tex]
Answer:
a) m = 330000 kg = 330 tons
b) H3 = 1.268 meters
Explanation:
Given:-
- The density of fresh-water, ρ = 1000 kg/m^3
- The base area of the rectangular prism boat, A = 550 m^2
- The height of the boat, H = 2.0 m ( empty )
- The bottom of boat barge is H1 = 0.45 m of the total height H under water. ( empty )
- The bottom of boat barge is H2 = 1.05 m of the total height H under water
Find:-
a) Find the mass of the coal in kilograms.
b) How far would the barge be submerged (in meters) if m; = 450000 kg of coal had been placed on the empty barge?
Solution:-
- We will consider the boat as our system with mass ( M ). The weight of the boat "Wb" acts downward while there is an upward force exerted by the body of water ( Volume ) displaced by the boat called buoyant force (Fb):
- We will apply the Newton's equilibrium condition on the boat:
Fnet = 0
Fb - Wb = 0
Fb = Wb
Where, the buoyant force (Fb) is proportional to the volume of fluid displaced ( V1 ). The expression of buoyant force (Fb) is given as:
Fb = ρ*V1*g
Where,
V1 : Volume displaced when the boat is empty and the barge of the boat is H1 = 0.45 m under the water:
V1 = A*H1
Hence,
Fb = ρ*A*g*H1
Therefore, the equilibrium equation becomes:
ρ*A*g*H1 = M*g
M = ρ*A*H1
- Similarly, apply the Newton's equilibrium condition on the boat + coal:
Fnet = 0
Fb - Wb - Wc = 0
Fb = Wb + Wc
Where, the buoyant force (Fb) is proportional to the volume of fluid displaced ( V2 ). The expression of buoyant force (Fb) is given as:
Fb = ρ*V2*g
Where,
V2 : Volume displaced when the boat is filled with coal and the barge of the boat is H2 = 1.05 m under the water:
V2 = A*H2
Hence,
Fb = ρ*A*g*H2
Therefore, the equilibrium equation becomes:
ρ*A*g*H2 = g*( M + m )
m+M = ρ*A*H2
m = ρ*A*H2 - ρ*A*H1
m = ρ*A*( H2 - H1 )
m = 1000*550*(1.05-0.45)
m = 330000 kg = 330 tons
- We will set the new depth of barge under water as H3, if we were to add a mass of coal m = 450,000 kg then what would be the new depth of coal H3.
- We will use the previously derived result:
m = ρ*A*( H3 - H1 )
H3 = m/ρ*A + H1
H3 = (450000 / 1000*550) + 0.45
H3 = 1.268 m
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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