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ira [324]
2 years ago
15

Find the time Δt it takes the magnetic field to drop to zero. Express your answer in terms of some or all of the quantities a, B

0, I, N, and R.
Chemistry
1 answer:
Mekhanik [1.2K]2 years ago
7 0
Δt= \frac{NB_0 \pi a^2 }{IR}

This because ε = dΦ/dt, and emf is a voltage so it can also be written as dΦ/dt = IR

Because ΔΦ= ABcosθ for each of the N loops and cosθ is just 1 here only A needs to be replaced, and it can be replaced with A=πr^2, which makes that part of the equation ΔΦ_m=N*B0*πa^2
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Perform the following conversion: 6.1 × 103 K (the surface temperature of the Sun) to °F and °C. Pay attention to the number of
Keith_Richards [23]

Answer:

\°C=5.8x10^3\°C

\°F=1.1x10^4\°F

Explanation:

Hello,

In this case, for the calculation of the temperature in degree Celsius we subtract 273.15 to the given temperature in kelvins:

\°C=6100-273.15\\\\\°C=5.8x10^3\°C

Next, by applying the following equation we compute it in degree Fahrenheit:

\°F=(5.8x10^{3}*9/5) + 32\\\\\°F=1.1x10^4\°F

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5 0
1 year ago
A solution is made by dissolving 58.125 g of sample of an unknown, nonelectrolyte compound in water. The mass of the solution is
e-lub [12.9K]

Answer:

molecular weight (Mb) = 0.42 g/mol

Explanation:

mass sample (solute) (wb) = 58.125 g

mass sln = 750.0 g = mass solute + mass solvent

∴ solute (b) unknown nonelectrolyte compound

∴ solvent (a): water

⇒ mb = mol solute/Kg solvent (nb/wa)

boiling point:

  • ΔT = K*mb = 100.220°C ≅ 373.22 K

∴ K water = 1.86 K.Kg/mol

⇒ Mb = ? (molecular weight) (wb/nb)

⇒ mb = ΔT / K

⇒ mb = (373.22 K) / (1.86 K.Kg/mol)

⇒ mb = 200.656 mol/Kg

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moles solute:

⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute

molecular weight:

⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol

8 0
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A person uses a bow to shoot an arrow at a target. In which diagram does the bow and arrow have the greatest amount of potential
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Answer: B

Explanation:

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You prepare a standard by weighing 10.751 mg of compound X into a 100 mL volumetric flask and making to volume. You further dilu
ArbitrLikvidat [17]

Answer:

0.12693 mg/L

Explanation:

First we <u>calculate the concentration of compound X in the standard prior to dilution</u>:

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Then we <u>calculate the concentration of compound X in the standard after dilution</u>:

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Now we calculate the<u> concentration of compound X in the sample</u>, using the <em>known concentration of standard and the given areas</em>:

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