A car moving at 60 mph slams on its brakes to stop before hitting a deer. Another identical car traveling at 60 mph slows to a s
1 answer:
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Answer:
the required minimum length of the attenuator is 3.71 cm
Explanation:
Given the data in the question;
we know that;
= 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
= 3 × 10¹⁰ / 2( 0.0711 )
= 3 × 10¹⁰ cm/s / 0.1422 cm
= 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)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;
20log₁₀ = -100dB
we substitute
20log₁₀ = -100dB
= 3.71 cm
Therefore, the required minimum length of the attenuator is 3.71 cm
To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:
Where,
F = Force
r = Radius
Replacing we have that,
The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore
Finally, angular acceleration is a result of the expression of torque by inertia, therefore
PART B)
The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians , therefore
Answer:
v=77.62 m/s
Explanation:
Given that
h= - 300 m
speed of the bird ,u= 5 m/s
Lets take Speed of the berry when it hit the ground = v m/s
we know that ,if object is moving upward
v² = u² - 2 g h
u=Initial speed
v=Final speed
h=Height
Now by putting the values
v² = u² - 2 g h
v² = 5² - 2 x 10 x (-300) ( take g = 10 m/s²)
v² =25 + 20 x 300
v² ==25 + 6000
v² =6025
v=77.62 m/s
Therefore the final speed of the berry will be 77.62 m/s.
