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A Venturi tube may be used as the inlet to an automobile carburetor. If an inlet pipe with a diameter of 2.0 cm diameter narrows to diameter of 1.0 cm, determine the pressure drop in the constricted section for an initial airflow of 3.0 m/s in the 2-cm section? (Assume air density is 1.25 kg/m

Answers

Answer 1

Answer:

The pressure drop is equal to 80.99 Pa

Given information:

d1 = 2 cm = 0.02 m

d2 = 1 cm = 0.01 m

v = 3 m/s

p = 1.25 kg/m^3

Here we use Bernoulli's principle for the Venturi Tube:

Calculation of pressure drop:

$P1 - P2 = ((v^2* p)/ 2)* ((A1^2/ A2^2)-1)\n\nP1 - P2 = \Delta P = ((v1^2* p)/ 2)* ((A1^2/ A2^2)-1)$

Now the following formula for area calculation should be used:

$A1 = (\pi* d1^2)/ 4 = (\pi* (0.02 m)^2)/ 4 = 0.00031 m^2\n\nA2 = (\pi* (0.01 m)^2)/ 4 = 0.000079 m^2\n\n\Delta P = ((3 m/s)^2 *1.25 kg/m^3)/ 2) * ((0.00031 m^2)^2/(0.000079 m^2)^2)-1)$

= 80.99

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Answer 2

Answer:

Answer:

the pressure drop is equal to 80.99 Pa

Explanation:

we have the following data:

d1 = 2 cm = 0.02 m

d2 = 1 cm = 0.01 m

v = 3 m/s

p = 1.25 kg/m^3

ΔP = ?

For the calculation of the pressure drop we will use Bernoulli's principle for the Venturi Tube:

P1 - P2 = ((v^2*p)/2)*((A1^2/A2^2)-1)

where A = area

P1 - P2 = ΔP = ((v1^2*p)/2)*((A1^2/A2^2)-1)

for the calculation of the areas we will use the following formula:

A1 = (pi*d1^2)/4 = (pi*(0.02 m)^2)/4 = 0.00031 m^2

A2 = (pi*(0.01 m)^2)/4 = 0.000079 m^2

ΔP = ((3 m/s)^2*1.25 kg/m^3)/2)*((0.00031 m^2)^2/(0.000079 m^2)^2)-1) = 80.99 N/m^2 = Pa

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Circuit A in a house has a voltage of 218 V and is limited by a 45-A circuit breaker. Circuit B is at 120.0 V and has a 25-A circuit breaker.What is the ratio of the maximum power delivered by circuit A to that delivered by circuit B?

Answers

Answer:

3.27

Explanation:

Electric Power: This can be defined as the rate at which electric energy is consumed. The unit of power is Watt (W).

Mathematically, electric power is represented as

P = VI ..................................... Equation 1.

Where P = power, V = voltage, I = Current.

For Circuit A,

P₁ = V₁I₁ ................................... Equation 2

Where P₁ = maximum power delivered by circuit A, V₁ = Voltage of circuit A, I₁ = circuit breaker rating of circuit A.

Given: V₁ = 218 V, I₁ = 45 A.

Substituting into equation 2

P₁ = 218×45

P₁ = 9810 W.

For Circuit B,

P₂ = V₂I₂............................. Equation 3

Where P₂ = maximum power delivered by the circuit B, V₂ = voltage of circuit B, I₂ = circuit breaker rating of circuit B

Given: V₂ = 120 V, I₂ = 25 A.

Substitute into equation 3

P₂ = 120(25)

P₂ = 3000 W.

Ratio of maximum power delivered by circuit A to that delivered by circuit B = 9810/3000

= 3.27.

Thus the ratio of maximum power delivered by circuit A to circuit B = 3.27

An electric current in a conductor varies with time according to the expression i(t) = 110 sin (120πt), where i is in amperes and t is in seconds. what is the total charge passing a given point in the conductor from t = 0 to t = 1/180 s?

Answers

As we know that current is defined as rate of flow of charge

$i = (dq)/(dt)$

so by rearranging the equation we can say

$q = \int i dt$

here we know that

$i(t) = 110 sin(120\pi t)$

here we will substitute it in the above equation

$q = \int 110 sin(120\pi t) dt$

$q = 110 [- (cos(120\pi t))/(120\pi)]$

now here limits of time is from t = 0 to t = 1/180s

so here it will be given as

$q = (110)/(120\pi)( -cos0 + cos((2\pi)/(3)))$

$q = 0.44 C$

so total charge flow will be 0.44 C

Answer:

The total charge passing a given point in the conductor is 0.438 C.

Explanation:

Given that,

The expression of current is

$i(t)=110\sin(120\pi t)$

$(dq(t))/(t)=110\sin(120\pi t)$

$dq(t)=110\sin(120\pi t)dt$ ....(I)

We need to calculate the total charge

On integrating both side of equation (I)

$\int_(0)^(q)dq(t)=\int_(0)^{(1)/(180)}110\sin(120\pi t)dt$

$q=110((-\cos(120\pi t))/(120\pi))_(0)^{(1)/(180)}$

$q=-(110)/(120\pi)(cos(120\pi((1)/(180)))-\cos120\pi(0))$

$q=-0.2918(-(1)/(2)-1)$

$q=0.438\ C$

Hence, The total charge passing a given point in the conductor is 0.438 C.

Solenoid A has total number of turns N length L and diameter D. Solenoid B has total number of turns 2N, length 2L and diameter 2D. Inductance of solenoid A is 8 times inductance of solenoid B
1/4 of inductance of solenoid B
same as inductance of solenoid B
1/8 of inductance of solenoid B
four times of inductance of solenoid B

Answers

Answer:

∴Inductance of solenoid A is $\frac18$ of inductance of solenoid B.

Explanation:

Inductance of a solenoid is

$L=N\frac\phi I$

$=N(B.A)/(I)$

$=N(\mu_0NI)/(l.I)A$

$=(\mu_0N^2A)/(l)$

$=(\mu_0N^2)/(l).\pi(\frac d2)^2$

$=\mu_0\pi(N^2d^2)/(4l)$

N= number of turns

$l$ = length of the solenoid

d= diameter of the solenoid

A=cross section area

B=magnetic induction

$\phi$ = magnetic flux

$I$ = Current

Given that, Solenoid A has total number of turns N, length L and diameter D

The inductance of solenoid A is

$=\mu_0\pi(N^2D^2)/(4L)$

Solenoid B has total number of turns 2N, length 2L and diameter 2D

The inductance of solenoid B is

$=\mu_0\pi((2N)^2(2D)^2)/(4.2L)$

$=\mu_0\pi(16 N^2D^2)/(4.2L)$

Therefore,

$\frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=(\mu_0\pi(N^2D^2)/(4L))/(\mu_0\pi(16 N^2D^2)/(4.2L))$

$\Rightarrow \frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=\frac18$

$\Rightarrow {\textrm{Inductance of A}}=\frac18* {\textrm{Inductance of B}}$

∴Inductance of solenoid A is $\frac18$ of inductance of solenoid B.

Hi there!

We can begin by calculating the inductance of a solenoid.

Recall:
$L = (\Phi _B)/(i)$

L = Inductance (H)
φ = Magnetic Flux (Wb)

i = Current (A)

We can solve for the inductance of a solenoid. We know that its magnetic field is equivalent to:
$B = \mu _0 (N)/(L)i$

And that the magnetic flux is equivalent to:
$\Phi _B = \int B \cdot dA = B \cdot A$

Thus, the magnetic flux is equivalent to:
$\Phi _B = \mu _0 (N)/(L)iA$

The area for the solenoid is the # of loops multiplied by the cross-section area, so:
$A_(total)= N * A$

$\Phi _B = \mu _0 (N^2)/(L)iA$

Using this equation, we can find how it would change if the given parameters are altered:
$\Phi_B ' = \mu_0 ((2N)^2)/(2L) i * 4A$

**The area will quadruple since a circle's area is 2-D, and you are doubling its diameter.

$\Phi'_B = (4)/(2) * 4(\mu_0 (N)/(L)iA) = 8\mu_0 (N)/(L)iA$

Thus, Solenoid B is 8 times as large as Solenoid A.

Solenoid A is 1/8 of the inductance of solenoid B.

A 140 W lightbulb emits 4% of its energy as electromagnetic radiation. What is the radiation pressure (in N/m2) on a perfectly absorbing sphere of radius 14 m that surrounds the bulb

Answers

Answer: 7.578x10^-12

Explanation:

First, we find the power:

Power P = 140x4/100 =5,6W

Distance r = 14m

Then,

Intensity I = P/4πr2

= 5.6/(4π x 14 x 14)

=. 2.27 x10^3 W/m2

Radiation pressure:

P(rad) = I/c =0.00227÷{3 x 10^8)

=7.578x10^-12 N/m2

Answer:

Pr=7.57*10^{11}Pa

Explanation:

We can solve this problem by taking into account the expression

$P_r=(IA)/(c)$

where I is the irradiance, c is the speed of light and A is the area.

We have that the power is 140W, but only 4% is electromagnetic energy, that is

$P=140W=140(J)/(s)\n0.4*140J=56J$

56J is the electromagnetic energy.

The area of the bulb is

$A_b=4\pi r^2=4\pi (14m)^2=2463m^2$

The radiation pressure is

$P_r=(56)/(2463m^2*3*10^8m/s^2)=7.57*10^(-11)Pa$

hope this helps!!

a weight is suspended from the ceiling by a spring (k = 20 ln/in) and is connected to the floor by a dashpot producing viscous damping. The damping force is 10 lb when the velocity of the dashpot plunger is 20 in/sec. The weight and plunger have W = 12 lb. What will be the frequency of the damped vibrations?