The formula for the cross sectional area of specimen at the middle is

A square loop of wire surrounds a solenoid. The side of the square is 0.1 m, while the radius of the solenoid is 0.025 m. The sq

Semmy [17]

Answer:

I=9.6×e^{-8} A

Explanation:

The magnetic field inside the solenoid.

B=I*500*muy0/0.3=2.1×e ^-3×I.

so the total flux go through the square loop.

B×π×r^2=I×2.1×e^-3π×0.025^2

=4.11×e^-6×I

we have that

(flux)'= -U

so differentiating flux we get

so the inducted emf in the loop.

U=4.11×e^{-6}×dI/dt=4.11×e^-6×0.7=2.9×e^-6 (V)

so, I=2.9×e^{-6}÷30

I=9.6×e^{-8} A

6 0

3 years ago

Define Plastic vs elastic deformation.

Snowcat [4.5K]

Answer:

Plastic deformation, irreversible or permanent. Deformation mode in which the material does not return to its original shape after removing the applied load. This happens because, in plastic deformation, the material undergoes irreversible thermodynamic changes by acquiring greater elastic potential energy.

Elastic deformation, reversible or non-permanent. the body regains its original shape by removing the force that causes the deformation. In this type of deformation, the solid, by varying its tension state and increasing its internal energy in the form of elastic potential energy, only goes through reversible thermodynamic changes.

3 0

3 years ago

Can I get an answer to this question please

crimeas [40]

Answer:

(i) 12 V in series with 18 Ω.

(ii) 0.4 A; 1.92 W

(iii) 1,152 J

(iv) 18Ω — maximum power transfer theorem

Explanation:

As seen by the load, the equivalent source impedance is ...

10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω

The open-circuit voltage seen by the load is ...

(36 V)(12/(24 +12)) = 12 V

The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.

__

The load current is ...

(12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current

The load power is ...

P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power

__

10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...

(600 s)(1.92 J/s) = 1,152 J

__

The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.

The power transferred to 18 Ω is ...

((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W

7 0

2 years ago

A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th

Taya2010 [7]

Answer:

(a) The heat transferred is 2552.64 kJ

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588 = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 = 2552.64 kJ

b) $\Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)$

Where:

$x_A$ and $x_B$ are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, $x_A$ = 248 /(248 + 42.8) = 0.83

$x_B$ = 42.8/(248 + 42.8) = 0.1472

∴ $\Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472)$ = 1066.0279 J/K

5 0

3 years ago

Note that common skills are listed toward the top, and less common skills are listed toward the bottom.

Sladkaya [172]

Answer:

BDEG

Explanation:

got it right on the test on edge because i used my b r a i n

5 0

3 years ago