Hi there,
for this question we have:
Signal 2.0 MHz = Emitted so we can call it
![f_e](https://tex.z-dn.net/?f=%20f_e)
and we need the Reflected =
![f_{r}](https://tex.z-dn.net/?f=%20f_%7Br%7D%20)
In this question, we have a source which goes to the heart and a reflected which comes back from the heart and we need the speed of the reflected.
So you should know that the speed of reflected is lower than the source(Emitted).
we also know: ΔBeat frequency(max) = 560 Hz =
![f_{b}](https://tex.z-dn.net/?f=%20f_%7Bb%7D%20)
so we have:
![f_{e}](https://tex.z-dn.net/?f=%20f_%7Be%7D%20)
-
![f_{r}](https://tex.z-dn.net/?f=%20f_%7Br%7D%20)
=
![f_{b}](https://tex.z-dn.net/?f=%20f_%7Bb%7D%20)
so frequency of Reflected is:
2.0 × 10^6 Hz - 560 Hz = 1.99 × 10^6 Hz =
![f_{r}](https://tex.z-dn.net/?f=%20f_%7Br%7D%20)
now you know that Lambda = v/f
so if we find the lambda with our Emitted then we can find v with the Reflected:
Lambda = 1540(m/s) / 2.0 × 10^6 Hz = 7.7 × 10^-4 m
=>
![v_{max}](https://tex.z-dn.net/?f=%20v_%7Bmax%7D%20)
= (lambda)(
=> 7.7 × 10^-4m (1.99 × 10^6Hz) = 1532 m/s
so the
![v_{max}](https://tex.z-dn.net/?f=%20v_%7Bmax%7D%20)
is equal to 1532 m/s :)))
This question is solved by two top teachers as fast as they could :))
I hope this is helpful
have a nice day
Answer:
3.83 m/s
Explanation:
Given that,
Distance covered by Jan, d = 4 miles
1 mile = 1609.34 m
4 miles = 6437.38 m
Time, t = 28 minutes = 1680 s
Jan's average speed,
v = d/t
![v=\dfrac{6437.38\ \text{m}}{1680\ \text{s}}\\\\v=3.83\ \text{m/s}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7B6437.38%5C%20%5Ctext%7Bm%7D%7D%7B1680%5C%20%5Ctext%7Bs%7D%7D%5C%5C%5C%5Cv%3D3.83%5C%20%5Ctext%7Bm%2Fs%7D)
Hence, the average velocity of Jan is 3.83 m/s.
Answer:
Vi = 32 [m/s]
Explanation:
In order to solve this problem we must use the following the two following kinematics equations.
![v_{f} =v_{i} - (a*t)\\](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bi%7D%20-%20%28a%2At%29%5C%5C)
The negative sign of the second term of the equation means that the velocity decreases, as indicated in the problem.
where:
Vf = final velocity = 8[m/s]
Vi = initial velocity [m/s]
a = acceleration = [m/s^2]
t = time = 5 [s]
Now replacing:
8 = Vi - 5*a
Vi = (8 + 5*a)
As we can see we have two unknowns the initial velocity and the acceleration, so we must use a second kinematics equation.
![v_{f}^{2} = v_{i}^{2} - (2*a*d)](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%20%3D%20v_%7Bi%7D%5E%7B2%7D%20-%20%282%2Aa%2Ad%29)
where:
d = distance = 100[m]
(8^2) = (8 + 5*a)^2 - (2*a*100)
64 = (64 + 80*a + 25*a^2) - 200*a
0 = 80*a - 200*a + 25*a^2
0 = - 120*a + 25*a^2
0 = 25*a(a - 4.8)
therefore:
a = 0 or a = 4.8 [m/s^2]
We choose the value of 4.8 as the acceleration value, since the zero value would not apply.
Returning to the first equation:
8 = Vi - (4.8*5)
Vi = 32 [m/s]
Rational expectations theory suggests that the speed of adjustment Purcell correction would be very quick.
<h3>What Is Rational Expectations Theory?</h3>
The rational expectations theory is a widely used concept and modeling technique in macroeconomics. Individuals make decisions based on three primary factors, according to the theory: their human rationality, the information available to them, and their past experiences.
The rational expectations hypothesis was originally suggested by John (Jack) Muth 1 (1961) to explain how the outcome of a given economic phenomena depends to a certain degree on what agents expect to happen.
- People who have rational expectations always learn from their mistakes.
- Forecasts are unbiased, and people make decisions based on all available information and economic theories.
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To learn more about Rational expectations theory from the given link
brainly.com/question/16479910
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a) 32.3 N
The force of gravity (also called weight) on an object is given by
W = mg
where
m is the mass of the object
g is the acceleration of gravity
For the ball in the problem,
m = 3.3 kg
g = 9.8 m/s^2
Substituting, we find the force of gravity on the ball:
![W=(3.3)(9.8)=32.3 N](https://tex.z-dn.net/?f=W%3D%283.3%29%289.8%29%3D32.3%20N)
b) 48.3 N
The force applied
![F_{app} = 36 N](https://tex.z-dn.net/?f=F_%7Bapp%7D%20%3D%2036%20N)
The ball is kicked with this force, so we can assume that the kick is horizontal.
This means that the applied force and the weight are perpendicular to each other. Therefore, we can find the net force by using Pythagorean's theorem:
![F=\sqrt{W^2+F_{app}^2}](https://tex.z-dn.net/?f=F%3D%5Csqrt%7BW%5E2%2BF_%7Bapp%7D%5E2%7D)
And substituting
W = 32.3 N
Fapp = 36 N
We find
![F=\sqrt{32.3^2+36^2}=48.3 N](https://tex.z-dn.net/?f=F%3D%5Csqrt%7B32.3%5E2%2B36%5E2%7D%3D48.3%20N)
c) ![14.6 m/s^2](https://tex.z-dn.net/?f=14.6%20m%2Fs%5E2)
The ball's acceleration can be found by using Newton's second law, which states that
F = ma
where
F is the net force on an object
m is its mass
a is its acceleration
For the ball in this problem,
m = 3.3 kg
F = 48.3 N
Solving the equation for a, we find
![a=\frac{F}{m}=\frac{48.3}{3.3}=14.6 m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7BF%7D%7Bm%7D%3D%5Cfrac%7B48.3%7D%7B3.3%7D%3D14.6%20m%2Fs%5E2)