emf generated by the coil is 1.57 V
Explanation:
Given details-
Number of turns of wire- 1000 turns
The diameter of the wire coil- 1 cm
Magnetic field (Initial)= 0.10 T
Magnetic Field (Final)=0.30 T
Time=10 ms
The orientation of the axis of the coil= parallel to the field.
We know that EMF of the coil is mathematically represented as –
E=N(ΔФ/Δt)
Where E= emf generated
ΔФ= change inmagnetic flux
Δt= change in time
N= no of turns*area of the coil
Substituting the values of the above variables
=1000*3.14*0.5*10-4
=.0785
E=0.0785(.2/10*10-3)
=1.57 V
Thus, the emf generated is 1.57 V
Answer:
16 minutes
Explanation:
This is an example of a class of problems in which two quantities start with different initial values and change at different rates. In such problems, the rates of change are generally ones that cause the values to converge.
The question usually asks when the values will be the same. The generic answer is, "when the difference in rates makes up the difference in initial values."
Here the tanks differ in initial fill height by 12 -8 = 4 ft. The rates of change differ by 0.5 -0.25 = 0.25 ft/min. The more filled tank is draining faster (important), so the fill heights will converge after ...
(4 ft)/(0.25 ft/min) = 16 min
The level in the two tanks will be the same after 16 minutes.
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<em>Additional comment</em>
The oil levels at that time will be 4 ft.
You can write two equations for height:
y = 12 -0.5x . . . . . . . height in feet after x minutes (tank A)
y = 8 -0.25x . . . . . . height in feet after x minutes (tank B)
These will be equal when ...
y = y
12 -0.5x = 8 -0.25x
4 = 0.25x . . . . . . . . . . add 0.5x -8
16 = x . . . . . . . . . . . . multiply by 4 . . . . time to equal height
The graph shows when the tanks will have equal heights and when they will be drained.
Answer:
Explanation:
There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.
We are given the following information:
y₁ = 3 m
y₂ = 3 m − 2.4 m = 0.6 m
t₂ − t₁ = 0.4 s
a = -9.8 m/s²
t₀ = 0 s
v₀ = 0 m/s
We need to find y₀.
Use a constant acceleration equation:
y = y₀ + v₀ t + ½ at²
Evaluated at point 1:
3 = y₀ + (0) t₁ + ½ (-9.8) t₁²
3 = y₀ − 4.9 t₁²
Evaluated at point 2:
0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²
0.6 = y₀ − 4.9 t₂²
Solve for y₀ in the first equation and substitute into the second:
y₀ = 3 + 4.9 t₁²
0.6 = (3 + 4.9 t₁²) − 4.9 t₂²
0 = 2.4 + 4.9 (t₁² − t₂²)
We know t₂ = t₁ + 0.4:
0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)
0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))
0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)
0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)
0 = 2.4 − 3.92 t₁ − 0.784
0 = 1.616 − 3.92 t₁
t₁ = 0.412
Now we can plug this into the original equation and find y₀:
3 = y₀ − 4.9 t₁²
3 = y₀ − 4.9 (0.412)²
3 = y₀ − 0.83
y₀ = 3.83
Rounded to two significant figures, the height of the tree is 3.8 meters.
Answer:
See explaination
Explanation:
See attachment for the detailed step by step solution of the given problem.
Answer:
Explanation:
(a) Given that 620g moisture and 330g decomposable organic matter in yard trimming is represented by C₁₂.₇₆H₂₁.₂₈O₉.₂₆N₀.₅₄
Given the atomic mass of Carbon C = 12, Hydrogen H = 1, Oxygen O = 16 and Nitrogen N = 14
1 mole of trimming = 12*12.76 + 1*21.28 + 16*9.26 + 14*0.54
= 153.12 + 21.28 + 148.16 + 7.56
= 330.12 g/mol
which means 1 kg of as received trimming has 330 g of decomposable that produce 1 mole of decomposable
The moles of methane produced will be given as
m = (4a + b -2c - 3d)/8
= (4*12.76 + 21.28 - 2*9.26 - 3*0.54)/8
= (51.04 + 21.28 - 18.52 - 1.62)/8
= 52.18/8
= 6.5225
(b) Volume of methane V is given as
V = (0.0224 m³ CH₄mol/CH₄) × (6.5225 mol CH₄/ kg)
= 0.1461 m³ CH₄/kg lawn trimmings
(c) Energy will be given as
CH₄Energy = 6.5225 mol of CH₄/kg × 890 kJ/mol
= 5805.025
≈ 5805 kJ/kg
