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Delicious77 [7]

2 years ago

12

A 10 m long high tension power line carries a current of 20 A perpendicular to Earth's magnetic field of 5.5 x10⁻⁵ T. What is th

e magnetic force experienced by the power line? ______ N *

1 answer:

den301095 [7]2 years ago

7 0

Answer:

Explanation:

Given that,

The length of tension wire is 10m

L = 10m

It caries a current of 10A

I = 10A

The current is perpendicular to the magnetic field

B = 5.5 × x10⁻⁵ T.

Then, θ = 90°

The magnetic force that act on the line is?

F = i(L × B)

L × B = LB Sinθ

Where

I is the current

L is the length of the wire

B is the magnetic field

θ is the angle between the field and the direction of the length.

Then,

F = I × L × B × Sinθ

F = 10 × 10 × 5.5 × x10⁻⁵ × Sin 90

F = 5.5 × x10⁻³ N

Then, the magnetic force experience on the line is 5.5 × x10⁻³N

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While ice skating, you unintentionally crash into a person. Your mass is 60 kg, and you are traveling east at 8.0 m/s with respe

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Answer:

6.18 m/s

Explanation:

Roller skate collision

The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;

x-axis component form (+x east);

$P_{Miy}$ + $p_{Piy}$ + $j_{y}$ = $P_{Mfy}$ + $P_{pfy}$

$m_{Mu_{Miy}+ m_{pu_{piy}}+0=(m_{M}+m_{p})V_{f} sin$ Ф

60 ·8 + 0 = (60 + 80) $V_{f}sin$ Ф

480 = 140 $V_{f} sin$ Ф................. (I)

y-axis component form (+y north);

$P_{Mix}$ + $p_{Pix}$ + $j_{x}$ = $P_{Mfx}$ + $P_{pfx}$

$m_{Mu_{Mix}+ m_{pu_{pix}}+0=(m_{M}+m_{p})V_{f} cos$ Ф

0 + 80.9 = (60 + 80) $V_{f}cos$ Ф

720= 140 $V_{f}cos$ Ф

140Vf= $\frac{720}{cos}$ Ф......................................(2)

Substituting (2) into (1) to give the angle;

480 = 720tan Ф

Ф = arctan(0.67) =33.69°.......................(3)

Evaluating (1) with (3) gives the velocity magnitude

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Vf=6.18 m/s

note 1:

This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.

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2 years ago

100-ft-long horizontal pipeline transporting benzene develops a leak 43 ft from the high-pressure end. The diameter of the leak

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Answer:

Explanation:

The mass flow rate of benzene from the leak in the pipeline containing benzene is:

$Q_m=AC_o\sqrt{2\rho g_cP_g}$

Here, $Q_m$ is the mass flow rate through the leak of the pipeline. $A$ is the area of the hole, $C_o$ is the discharge rate, $\rho$ is the fluid density, $g_c$ is the gravitational constant and $P_g$ is the constant gauge pressure within the process unit.

The diametre of the leak (d) is 0.1 in. Convert from in to ft.

$d=(0.1 in)(\frac{1ft}{12in})\\=8.33\times 10^{-3}ft$

Calculate the area (A) of the hole. The area of the hole is.

$A=\frac{\pi d^2}{4}$

Substitute 3.14 for $\pi$ and $8.33\times 10^{-3}ft$ for d and calculate A.

$A=\frac{\pi d^2}{4}\\\\\frac{(3.14)(8.33\times 10^{-3})^2}{4}\\\\5.45\times 10^{-5}ft^2$

The specific gravity of benzene is 0.8794. Specific gravity is the ratio of th density of a substance to the density of a reference substance.

Specific gravity of benzene = density of benzenee/denity of reference substance

Rewrite the expression in terms of density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

Take the reference substance as water. Density of water is $62.4\frac{Ib_m}{ft^3}$ . Calculate density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

$=(0.8794)(62.4\frac{Ib_m}{ft^3})\\\\54.9\frac{Ib_m}{ft^3}$

Calculate the pressure at the point of leak. The pressure is the average of the pressure of the high and low pressure end. Write the expression to calculate the average pressure.

Upstream x distance from upstream pressure end

$P_g$ =+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

Calculate the distance from the downstream pressure end. The distance from upstream pressure end is 43 ft. Total of the pipe is 100 ft.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

The distance from upstream pressure end is 43 ft. Total length of the pipe is 100 ft. Substitute the values in the equation.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

= 100ft - 43ft = 57 ft

Substitute 50 psig for upstream, 43 ft fr distance from the upstream pressure end, 40 psig for downstream pressure, 57 ft for distance from the downstream pressure end, and 100 ft for the total length of the horizontal pipeline and calculate $P_g$ .

Upstream x distance from upstream pressure end

$P_g$ =+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

$=\frac{(50psig\times 43ft)+(40psig \times 57ft)}{100ft}\\\\=44.3psig$

Convert the pressure from psig to $Ib_f/ft^2$

$P_g=(44.3psig)(\frac{1\frac{Ib_f}{ft^2}}{1psig})(144\frac{in^2}{ft^2})\\\\=6,379.2\frac{Ib_f}{ft^2}$

The leak is like a sharp orifice. Take the value of the discharge coefficient as 0.61.

Substitute $5.45\times 10^{-5}ft^2$ for A. 0.61 for $C_o$ , $54.9\frac{Ib_m}{ft^3}$ for $\rho$ , $32.17\frac{ft.Ib_m}{Ib_f.s^2}$ for $g_c$ , and $6,379.2\frac{Ib_f}{ft^2}$ for $P_g$ and calculate $Q_m$

$Q_m=AC_o\sqrt{2\rho g_cP_g}\\\\=(5.45\times 10^{-5}ft^2)(0.61)\sqrt{2(54.9\frac{Ib_m}{ft^3})(32.17\frac{ft.Ib_m}{Ib_f.s^2})(6,379.2\frac{Ib_f}{ft^2})}\\\\(3.3245\times 10^{-5}ft^2)\sqrt{22,533,031.21\frac{Ib^2_m}{ft^4.s^2}}\\\\=0.158\frac{Ib_m}{s}$

The mass flow rate of benzene through the leak in the pipeline is $0.158\frac{Ib_m}{s}$

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