Answer:
λ₁ = 2.50 10⁻² m, λ₂ = 1.66 10⁻² m
Explanation:
Microwave communication is very efficient because it does not have atmospheric interference, for which it is widely used and has been regulated to avoid interference, the ku band is in the range between 12 and 18 GHz.
Let's calculate the wavelength for the two extreme frequencies of this band
wavelength and frequency are related
c = λ f
λ = c / f
f₁ = 12 GHz = 12 10⁹ Hz
λ₁ = 3 10⁸ /12 10⁹
λ₁ = 2.50 10⁻² m
f₂ = 18 GHz = 18 10⁹ Hz
λ₂ = 3 10⁸ /18 10⁹
λ₂ = 1.66 10⁻² m
Unfortunately in your exercise the specific frequency is not fired, for significant figures they must be the same number as the figures of the frequency, in general the frequency has 3 or 4 significant figures
Answer:
5694000 min
Explanation:
Let's suppose the average American watches 4 hours of TV every day. First, we will calculate how many minutes they watch per day. We will use the conversion factor 1 h = 60 min.
(4 h/day) × (60 min/1 h) = 240 min/day
They watch 240 minutes of TV per day. Now, let's calculate how many minutes they watch per year. We will use the conversion factor 1 year = 365 day.
240 min/day × (365 day/year) = 87600 min/year
They watch 87600 min/year. Finally, let's calculate how many minutes they spend watching TV in 65 years.
87600 min/year × 65 year = 5694000 min
Answer:
v_s = 34.269 m / s
Explanation:
This is a Doppler effect exercise, in this case the observer is fixed and the sound source is moving.
f ’= f 
where the negative sign is used for when the source approaches the observer and the positive sign for when the source moves away from the observer
In this case when f ’= 5500 Hz approaches and when f’ = 4500 Hz moves away, let's write the two expressions together
5500 = f (
)
4500 = f ( 
)
let's solve these two equations
1.222 (v-v_s) = v + v_s
v_s (1+ 1.22) = v (1.222 -1)
v_s = v 
the speed of sound in air is v = 343 m / s
v_s = 343 0.09990
v_s = 34.269 m / s
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
I don’t know the first one but the second one is light and the third one is a satellite
