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3 years ago

Using a slotted line, the following results were obtained:

Physics

1 answer:

MAVERICK [17]3 years ago

6 0

Answer:

$Z_L=67 - j16.4$ Ω

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

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

$Z_L=67 - j16.4$ Ω

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The joule and the kilowatt-hour are both units of energy. 15 kw · h is equivalent to how many joules? answer in units of j.

choli [55]

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

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

3 0

3 years ago

What happens to thermal energy when Kinetic energy is Increased/Decreased?

Elis [28]

Answer:If kinetic energy increases, so does the thermal energy, and vice versa.

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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:

$a_{r} = 4\pi^{2}\cdot f^{2}\cdot R$ (1)

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)