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Anna71 [15]
3 years ago
14

What is the minimum frequency with which a 200-turn, flat coil of cross sectional area 300 cm2 can be rotated in a uniform 30-mT

magnetic field if the maximum value of the induced emf is to equal 8.0 V
Physics
1 answer:
sergiy2304 [10]3 years ago
8 0

Answer:

The minimum frequency of the coil is 7.1 Hz

Explanation:

Given;

number of turns, N = 200 turns

cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²

magnitude of magnetic field strength, B = 30 x 10⁻³ T

maximum value of the induced emf, E = 8 V

Maximum induced emf is given as;

E = NBAω

where

ω is angular velocity (ω = 2πf)

E = NBA2πf

where;

f is the minimum frequency, measured in hertz (Hz)

f = E / (NBA2π)

f = 8 / (200 x 30 x 10⁻³  x 300 x 10⁻⁴ x 2 x 3.142)

f = 7.073 Hz

f = 7.1 Hz

Therefore, the minimum frequency of the coil is 7.1 Hz

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How was the gravitational constant G first determined
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In the diagram below, what is the property of the wave indicated by the letter A? a.Crest
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Phosphorous reacts with chlorine gas to produce phosphorous pentachloride. Calculate the mass of product produced when 25.0 g of
MariettaO [177]

Answer : The mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.

Explanation : Given,

Mass of P = 25 g

Mass of Cl_2 = 25 g

Molar mass of P = 30.97 g/mole

Molar mass of Cl_2 = 71 g/mole

Molar mass of PCl_5 = 208.24 g/mole

First we have to calculate the moles of P and Cl_2.

\text{Moles of }P=\frac{\text{Mass of }P}{\text{Molar mass of }P}=\frac{25g}{30.97g/mole}=0.807moles

\text{Moles of }Cl_2=\frac{\text{Mass of }Cl_2}{\text{Molar mass of }Cl_2}=\frac{25g}{71g/mole}=0.352moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2P+5Cl_2\rightarrow 2PCl_5

From the balanced reaction we conclude that

As, 5 moles of Cl_2 react with 2 moles of P

So, 0.352 moles of Cl_2 react with \frac{2}{5}\times 0.352=0.1408 moles of P

That means, in the given balanced reaction, Cl_2 is a limiting reagent and it limits the formation of products and P is an excess reagent because the given moles are more than the required moles.

Now we have to calculate the moles of PCl_5.

As, 5 moles of Cl_2 react with 2 moles of PCl_5

So, 0.352 moles of Cl_2 react with \frac{2}{5}\times 0.352=0.1408 moles of PCl_5

Now we have to calculate the mass of PCl_5.

\text{Mass of }PCl_5=\text{Moles of }PCl_5\times \text{Molar mass of }PCl_5

\text{Mass of }PCl_5=(0.1408mole)\times (208.24g/mole)=29.32g

Now we have to calculate the mass of product produced (actual yield).

\%\text{ yield of }PCl_5=\frac{\text{Actual yield of }PCl_5}{\text{Theoretical yield of }PCl_5}\times 100

70.5=\frac{\text{Actual yield of }PCl_5}{29.32g}\times 100

\text{Actual yield of }PCl_5=20.67g

Therefore, the mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.

3 0
4 years ago
You throw a 3.00 N rock vertically into the air from ground level. You observe that when it is 16.0 m above the ground, it is tr
Morgarella [4.7K]

Answer:

The final speed of the stone as it lift the ground is 23.86 m/s.            

Explanation:

Given that,

Force acting on the rock, F = 3 N

Distance, d = 16 m

Initial speed of the stone, u = 22 m/s

We need to find the rock's speed just as it left the ground. It can be calculated using work energy theorem as :

W=\Delta E\\\\W=\dfrac{1}{2}m(v^2-u^2)\\\\Fd=\dfrac{1}{2}m(v^2-u^2)\\\\v^2=\dfrac{2Fd}{m}+u^2\\\\v^2=\dfrac{2mgd}{m}+u^2\\\\v^2=2\times 9.8\times 16+(16)^2\\\\v=23.86\ m/s

So, the final speed of the stone as it lift the ground is 23.86 m/s.                                                    

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