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11111nata11111 [884]
3 years ago
14

What is the answer to this question

Physics
1 answer:
deff fn [24]3 years ago
3 0

Answer:75%effecient

Explanation:

6000-1500=4500

6000=100%

4500=?

(4500×100)÷6000

=75%

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A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc
adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

f_{c_1 = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

f_{c_1 = 3 × 10¹⁰ / 2( 0.0711  )

f_{c_1 = 3 × 10¹⁰ cm/s / 0.1422  cm

f_{c_1 =  21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( (f_{c_1 / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀e^{-\alpha _1l_{min = -100dB

we substitute

20log₁₀e^{-(310.7np/m)l_{min = -100dB

l_{min = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0
3 years ago
A body has masses of 0.013kg and 0.012kg in oil and water respectively, if the relative density of oil is 0.875, calculate the m
konstantin123 [22]

Answer:

the mass of the body is 0.02 kg.

Explanation:

Given;

relative density of the oil, \gamma _0 = 0.875

mass of the object in oil, M_o = 0.013 kg

mass of the object in water, M_w = 0.012 kg

let the mass of the object in air = M_a

weight of the oil, W_0 = M_a - 0.013

weight of the water, W_w = M_a - 0.012

The relative density of the oil is given as;

\gamma_0 = \frac{density \ of \ oil }{density \ of \ water} = \frac{W_0}{W_w} = \frac{M_a -0.013}{M_a -0.012} \\\\0.875 = \frac{M_a -0.013}{M_a -0.012}\\\\0.875(M_a - 0.012) = M_a - 0.013\\\\0.875M_a - 0.0105 = M_a -0.013\\\\0.875M_a - M_a = 0.0105 - 0.013\\\\-0.125 M_a = -0.0025\\\\M_a = \frac{0.0025}{0.125} \\\\M_a = 0.02 \ kg

Therefore, the mass of the body is 0.02 kg.

6 0
3 years ago
A slinky is stretched by two forces. When the forces are removed, the slinky returns to its original length. The slinky has been
e-lub [12.9K]

Answer: <u>elastically</u> deformed or <u>non-permanently</u> deformed

Explanation:

According to classical mechanics, there are two types of deformations:

-Plastic deformation (also called irreversible or permanent deformation), in which the material does not return to its original form after removing the applied force, therefore it is said that the material was permanently deformed.

This is because the material undergoes irreversible thermodynamic changes while it is subjected to the applied forces.

-Elastic deformation (also called reversible or non-permanent deformation), in which the material returns to its original shape after removing the applied force that caused the deformation.

In this case t<u>he material also undergoes thermodynamic changes, but these are reversible, causing an increase in its internal energy by transforming it into elastic potential energy.</u>

<u />

Therefore, the situation described in the question is related to elastic deformation.

8 0
3 years ago
In the eastern US, hurricanesusually form over the warm waters of the Caribbean Sea and travel north. When they are over the col
garik1379 [7]

Answer:

Less powerful

Explanation:

Hurricanes rely on warm water. It sucks heat energy from the water to use for fuel. Warmer water means more moisture, which also mean a bigger and/or stronger hurricane. The North Atlantic is definitely much colder than the Caribbean so the hurricane will not have much fuel.

Have a great day!

8 0
3 years ago
The position of an electron is given by , with t in seconds and in meters. At t = 3.99 s, what are (a) the x-component, (b) the
egoroff_w [7]

Answer:

A. Vx = 3.63 m/s

B. Vy = -45.73 m/s

C. |V| = 45.87 m/s

D. θ = -85.46°

Explanation:

Given that position, r, is given as:

r = 3.63tˆi − 5.73t^2ˆj + 8.16ˆk

Velocity is the derivative of position, r:

V = dr/dt = 3.63 - 11.46t^j

A. x component of velocity, Vx = 3.63 m/s

B. y component of velocity, Vy = -11.46t

t = 3.99 secs,

Vy = - 11.46 * 3.99 = -45.73 m/s

C. Magnitude of velocity, |V| = √[(-45.73)² + 3.63²]

|V| = √(2091.2329 + 13.1769)

|V| = √(2104.4098)

|V| = 45.87 m/s

D. Angle of the velocity relative to the x axis, θ is given as:

tanθ = Vy/Vx

tanθ = -45.73/3.63

tanθ = -12.6

θ = -85.46°

7 0
4 years ago
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