TRUE of FALSE: The human body responds to stressors by activating the nervous

What happened on may 24 2002 in history

Leviafan [203]

Russia and the United States sign the Moscow Treaty.

5 0

3 years ago

The inductance of a solenoid with 450 turns and a length of 24 cm is 7.3 mH. (a) What is the cross-sectional area of the solenoi

Novay_Z [31]

Answer: 0.43 V

Explanation:

L = [μ(0) * N² * A] / l

Where

L = Inductance of the solenoid

N = the number of turns in the solenoid

A = cross sectional area of the solenoid

l = length of the solenoid

7.3*10^-3 = [4π*10^-7 * 450² * A] / 0.24

1.752*10^-3 = 4π*10^-7 * 202500 * A

1.752*10^-3 = 0.255 * A

A = 1.752*10^-3 / 0.255

A = 0.00687 m²

A = 6.87*10^-3 m²

emf = -N(ΔΦ/Δt).........1

L = N(ΔΦ/ΔI) so that,

N*ΔΦ = ΔI*L

Substituting this in eqn 1, we have

emf = - ΔI*L / Δt

emf = - [(0 - 3.2) * 7.3*10^-3] / 55*10^-3

emf = 0.0234 / 0.055

emf = 0.43 V

6 0

3 years ago

The velocity profile in fully developed laminar flow in a circular pipe of inner radius R 5 2 cm, in m/s, is given by u(r) 5 4(1

xxMikexx [17]

The question is not clear and the complete clear question is;

The velocity profile in fully developed laminar flow In a circular pipe of inner radius R = 2 cm, in m/s, is given By u(r) = 4(1 - r²/R²). Determine the average and maximum Velocities in the pipe and the volume flow rate.

Answer:

A) V_max = 4 m/s

B) V_avg = 2 m/s

C) Flow rate = 0.00251 m³/s

Explanation:

A) We are given that;

u(r) = 4(1 - (r²/R²))

To obtain the maximum velocity, let's apply the maximum condition for a single-variable continual real valued problem to obtain;

(d/dr)(u(r)) = 0

Thus,

(d/dr)•4(1 - (r²/R²)) = 0

4(d/dr)(1 - (r²/R²)) = 0

If we differentiate, we have;

4(0 - (2r/R²)) = 0

-8r/R² = 0

Thus, r = 0 and with that, the maximum velocity is at the centre of the pipe.

Thus, for maximum velocity, let's put 0 for r in the U(r) function.

Thus,

V_max = 4(1 - 0²/R²) = 4 - 0 = 4 m/s

B) Average velocity is given by;

V_avg = V_max/2

V_avg = 4/2 = 2 m/s

C) the flow can be calculated from;

Flow rate ΔV = A•V_avg

A is area = πr²

From question, r = 2cm = 0.02m

A = π x 0.02²

Hence,

ΔV = π x 0.02² x 2 = 0.00251 m³/s

8 0

3 years ago

The stopcock connecting a 2.14 L bulb containing oxygen gas at a pressure of 8.19 atm, and a 9.84 L bulb containing krypton gas

marshall27 [118]

Answer : The final pressure of the system in atm is, 3.64 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$