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2 years ago

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8) A flat circular loop having one turn and radius 5.0 cm is positioned with its plane perpendicular to a uniform 0.60-T magneti

c field. The area of the loop is suddenly reduced to essentially zero in 0.50 ms. What emf is induced in the loop

1 answer:

Marrrta [24]2 years ago

7 0

Answer:

EMF induced in the loop is 9.4 V

Explanation:

As we know that initial magnetic flux of the loop is given as

$\phi_1 = B.A$

$\phi_1 = (0.60)(\pi (0.05)^2)$

$\phi_1 = 4.7 \times 10^{-3} Wb$

As soon as the area of the loop becomes zero the final magnetic flux of the loop is ZERO

Now as per faraday's law of electromagnetic induction the EMF is induced due to rate of change in magnetic flux

so we will have

$EMF = \frac{\Delta \phi}{\Delta t}$

so we will have

$EMF = \frac{4.7 \times 10^{-3} - 0}{0.50 \times 10^{-3}}$

$EMF = 9.4 V$

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A cylindrical rod of steel (E = 207 GPa, 30 × 10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a

Yanka [14]

Answer:

Diameter of the cylinder will be $d=2.998\times 10^4m$

Explanation:

We have given young's modulus of steel $E=207GPa=207\times 10^9Pa$

Change in length $\Delta l=0.38mm$

Length of rod $l=500mm$

Load F = 11100 KN

Strain is given by $strain=\frac{\Delta l}{l}=\frac{0.38}{500}=7.6\times 10^{-4}$

We know that young's modulus $E=\frac{stress}{strain}$

So $207\times 10^9=\frac{stress}{7.6\times 10^{-4}}$

$stress=1573.2\times 10^{-5}N/m^2$

We know that stress $=\frac{force}{artea }$

So $1573.2\times 10^{-5}=\frac{11100\times 1000}{area}$

$area=7.055\times 10^{8}m^2$

So $\frac{\pi }{4}d^2=7.055\times 10^{8}$

$d=2.998\times 10^4m$

6 0

2 years ago

You are stranded in a blizzard. You need water to drink to drink,and you're trying to stay warm.should the melt the snow and dri

Crazy boy [7]

It would be a really bad idea to eat the snow because you obviously are trying to stay warm right? Well, the best thing to do is melt the snow. However, the process of melting the snow would have a few complications as well. But yes, the latter idea (drinking the snow) is a better idea (not the best).

7 0

2 years ago

A bowling ball starts from rest and moves 300 m down a long, downwardly angled track in 22.4 sec.a. What is its speed at the end

Sergio [31]

Answer:

The speed at the end of the track = 27 m/s

The acceleration = 1.2 m/s²

Please find the Δx vs Δt, v vs Δt, a vs Δt

Explanation:

We have;

x = u·t + 1/2·a·t²

Where;

x = The distance = 300 m

u = The initial velocity = 0 m/s (Ball at rest)

t = The time taken = 22.4 s

Therefore;

300 = 0 + 1/2×a×22.4²

a = 2×300/22.4² = 1.19579 ≈ 1.2 m/s²

v = u + a×t

∴ v = 0 + 1.2 × 22.4 = 26.88 ≈ 27 m/s

Part of the table of values is as follows;

t, x, v

0, 0, 0

0.4, 0.095663, 0.478316

0.8, 0.382653, 0.956632

1.2, 0.860969, 1.434948

1.6, 1.530611, 1.913264

2, 2.39158, 2.39158

2.4, 3.443875, 2.869896

2.8, 4.687497, 3.348212

3.2, 6.122445, 3.826528

3.6, 7.748719, 4.304844

4 0

2 years ago

Suppose that we use a heater to boil liquid nitrogen (N2 molecules). 4480 J of heat turns 20 g of liquid nitrogen into gas. Note

love history [14]

Answer:

The energy is $9.4\times10^{-21}\ J$

(a) is correct option

Explanation:

Given that,

Energy = 4480 j

Weight of nitrogen = 20 g

Boil temperature = 77 K

Pressure = 1 atm

We need to calculate the internal energy

Using first law of thermodynamics

$Q=\Delta U+W$

$Q=\Delta U+nRT$

Put the value into the formula

$4480=\Delta U+\dfrac{20}{28}\times8.314\times77$

$\Delta U=4480-\dfrac{20}{28}\times8.314\times77$

$\Delta U=4022.73\ J$

We need to calculate the number of molecules in 20 g N₂

Using formula of number of molecules

$N=n\times \text{Avogadro number}$

Put the value into the formula

$N=\dfrac{20}{28}\times6.02\times10^{23}$

$N=4.3\times10^{23}$

We need to calculate the energy

Using formula of energy

$E=\dfrac{\Delta U}{N}$

Put the value into the formula

$E=\dfrac{4022.73}{4.3\times10^{23}}$

$E=9.4\times10^{-21}\ J$

Hence, The energy is $9.4\times10^{-21}\ J$

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2 years ago

A planet has a surface temperature of 95.0 K and an atmospheric pressure of 1.60 atm. If 4.87 L of atmosphere has a mass of 28.6

mr Goodwill [35]

Answer:

M=28.88 gm/mol

Explanation:

Given that

T= 95 K

P= 1.6 atm

V= 4.87 L

m = 28.6 g

R=0.08206L atm .mol .K

We know that gas equation for ideal gas

P V = n R T

P=Pressure , V=Volume ,n=Moles,T= Temperature ,R=gas constant

Now by putting the values

P V = n R T

1.6 x 4.87 = n x 0.08206 x 95

n=0.99 moles

We know that number of moles given as

$n=\dfrac{m}{M}$

M=Molar mass

$n=\dfrac{m}{M}$

$0.99=\dfrac{28.6}{M}$

M=28.88 gm/mol

3 0

2 years ago