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bonufazy [111]
2 years ago
7

A magnetic field is directed perpendicular to the plane of a 0.15-m × 0.30-m rectangular coil consisting of 240 loops of wire. T

o induce an average emf of –2.5 V in the coil, the magnetic field is increased from 0.1 T to 1.8 T during a time interval Δt. Determine Δt.
Physics
1 answer:
ValentinkaMS [17]2 years ago
6 0

Answer:

7.344 s

Explanation:

A = 0.15 x 0.3 m^2 = 0.045 m^2

N = 240

e = - 2.5 v

B1 = 0.1 T

B2 = 1.8 T

ΔB = B2 - B1 = 1.8 - 0.1 = 1.7 T

Δt = ?

e = - dФ/dt

e = - N x A x ΔB/Δt

- 2.5 = - 240 x 0.045 x 1.7 / Δt

2.5 = 18.36 / Δt

Δt = 7.344 s

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Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

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Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

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          The altitude of container in Denver is  d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m

           The surface area of the container lid is A = 0.0155m^2

           The altitude of container in New Orleans  is sea-level

           The air pressure in Denver is  P_D = 79000 \ Pa

            The air pressure in new Orleans is P_{ro} = 100250 \ Pa

Generally force is mathematically represented as

            F_{No} = \Delta P A

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               P_{sea} = 101000 Pa

So the \Delta P which is the difference in pressure within and outside the container is  

           \Delta P = P_{No} - \frac{P_{sea}}{2}

Therefore

                F_{No} =   A [P_{No} - \frac{P_{sea}}{2}]

Now substituting values

                F_{No} =   0.0155 [100250 - \frac{101000}{2}]

                       F_{No}= 7771.125 \ N

The force to remove the lid in Denver is  

           F_p = \Delta P_d A

So we are told the pressure inside is  is half the pressure the at sea level so the  the pressure acting on the container would

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               P_{sea} = 101000 Pa    

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              P_D = \rho g h

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                        \Delta P_d = 1.44*10^{8}Pa

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                  = 1.44*10^8 * 0.0155

              F_p = 2.2*10^{6} N

               

                 

             

             

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