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Crank
3 years ago
9

Using a slotted line, the following results were obtained:

Physics
1 answer:
MAVERICK [17]3 years ago
6 0

Answer:

Z_L=67 - j16.4Ω

Explanation:

Given that

d(min,0)= 4 cm

d(min,1)= 14 cm

Voltage standingwave ratio = 1.5

Zo = 50 Ω

We know that

d(min,1) - d(min,0) = λ/2

Now by putting the values

14 - 4 = λ/2

λ = 20 cm

We also know that

β=2π/λ

β=2π/0.2 = 10π rad/m

So we can say that

θr= 2β d(min,n) - (2 n + 1)π rad

θr=2×10π ×0.04 −π = -0.2 π rad

We know that

π rad = 180 °

θr= = -0.2 π rad= -36 °

We know that

\Gamma =\dfrac{S-1}{S+1}

Here S= 1.5

\Gamma =\dfrac{S-1}{S+1}

\Gamma =\dfrac{1.5-1}{1.5+1}

\Gamma =0.2

Z_L=Z_o\dfrac{1+\Gamma }{1-\Gamma }

by putting the values

Z_L=50\times \dfrac{1+0.2e^{-j36^{\circ}} }{1-0.2e^{-j36^{\circ}} }

Z_L=67 - j16.4Ω

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The solution for the problem is:

1 Watt = 1 Joule per second 
1 Watt*second = 1 Joule 

a Kilowatt is 1,000 Watts 
an hour is 60 seconds times 60 minutes or 3,600 seconds 
a Kilowatt * hour is 1,000 Watts in 3,600 seconds 

15 W*h = 15,000 Watt*hour = 15,000 Watt * 3,600 seconds = 54,000,000 Watt*second 

54,000,000 Watt*second = ? Joules 
54,000,000 Joules / second = 54,000,000 Watts

3 0
3 years ago
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

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3 years ago
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3 0
3 years ago
One end of a string is fixed. An object attached to the other end moves on a horizontal plane with uniform circular motion of ra
sveticcg [70]

Answer:

If both the radius and frequency are doubled, then the tension is increased 8 times.

Explanation:

The radial acceleration (a_{r}), measured in meters per square second, experimented by the moving end of the string is determined by the following kinematic formula:

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Where:

f - Frequency, measured in hertz.

R - Radius of rotation, measured in meters.

From Second Newton's Law, the centripetal acceleration is due to the existence of tension (T), measured in newtons, through the string, then we derive the following model:

\Sigma F = T = m\cdot a_{r} (2)

Where m is the mass of the object, measured in kilograms.

By applying (1) in (2), we have the following formula:

T = 4\pi^{2}\cdot m\cdot f^{2}\cdot R (3)

From where we conclude that tension is directly proportional to the radius and the square of frequency. Then, if radius and frequency are doubled, then the ratio between tensions is:

\frac{T_{2}}{T_{1}} = \left(\frac{f_{2}}{f_{1}} \right)^{2}\cdot \left(\frac{R_{2}}{R_{1}} \right) (4)

\frac{T_{2}}{T_{1}} = 4\cdot 2

\frac{T_{2}}{T_{1}} = 8

If both the radius and frequency are doubled, then the tension is increased 8 times.

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3 years ago
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