<h2>Answer: True
</h2>
The <u>Doppler effect</u> refers to the change in a wave perceived frequency when the emitter of the waves, and the receiver (or observer in the case of light) move relative to each other.
In other words, it is the variation of the frequency of a wave due to the relative movement of the source of the wave with respect to its receiver.
It should be noted that this effect bears its name in honor of the Austrian physicist <u>Christian Andreas Doppler</u>, who in 1842 proposed the existence of this effect for the case of light in the stars. Another important aspect is that the effect occurs in all waves (including light and sound). However, it is more noticeable to humans with sound waves.
Answer:
The value is ![v = 2.3359 *10^{4} \ m/s](https://tex.z-dn.net/?f=v%20%20%3D%20%202.3359%20%2A10%5E%7B4%7D%20%5C%20m%2Fs)
Explanation:
From the question we are told that
The initial speed is ![u = 2.05 *10^{4} \ m/s](https://tex.z-dn.net/?f=u%20%3D%20%202.05%20%2A10%5E%7B4%7D%20%5C%20%20m%2Fs)
Generally the total energy possessed by the space probe when on earth is mathematically represented as
![T__{E}} = KE__{i}} + KE__{e}}](https://tex.z-dn.net/?f=T__%7BE%7D%7D%20%3D%20%20KE__%7Bi%7D%7D%20%2B%20%20KE__%7Be%7D%7D)
Here
is the kinetic energy of the space probe due to its initial speed which is mathematically represented as
=>
=> ![KE_i = 2.101 *10^{8} \ \ m \ \ J](https://tex.z-dn.net/?f=KE_i%20%3D%20%202.101%20%2A10%5E%7B8%7D%20%5C%20%5C%20m%20%5C%20%5C%20J)
And
is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as
![KE_e = \frac{1}{2} * m * v_e^2](https://tex.z-dn.net/?f=KE_e%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20m%20%2A%20%20v_e%5E2)
Here
is the escape velocity from earth which has a value ![v_e = 11.2 *10^{3} \ m/s](https://tex.z-dn.net/?f=v_e%20%3D%20%2011.2%20%2A10%5E%7B3%7D%20%5C%20%20m%2Fs)
=> ![KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2](https://tex.z-dn.net/?f=KE_e%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20m%20%2A%20%20%2811.3%20%2A10%5E%7B3%7D%29%5E2)
=> ![KE_e = 6.272 *10^{7} \ \ m \ \ J](https://tex.z-dn.net/?f=KE_e%20%3D%20%206.272%20%2A10%5E%7B7%7D%20%5C%20%20%5C%20%20m%20%20%5C%20%5C%20%20%20J)
Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as
![KE_p = \frac{1}{2} * m * v^2](https://tex.z-dn.net/?f=KE_p%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20m%20%2A%20%20v%5E2)
Generally from the law energy conservation we have that
So
![2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2](https://tex.z-dn.net/?f=2.101%20%2A10%5E%7B8%7D%20%20m%20%20%2B%20%206.272%20%2A10%5E%7B7%7D%20%20m%20%20%3D%20%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20m%20%2A%20%20v%5E2)
=> ![5.4564 *10^{8} = v^2](https://tex.z-dn.net/?f=5.4564%20%2A10%5E%7B8%7D%20%3D%20%20%20v%5E2)
=> ![v = \sqrt{5.4564 *10^{8}}](https://tex.z-dn.net/?f=v%20%3D%20%20%5Csqrt%7B5.4564%20%2A10%5E%7B8%7D%7D)
=> ![v = 2.3359 *10^{4} \ m/s](https://tex.z-dn.net/?f=v%20%20%3D%20%202.3359%20%2A10%5E%7B4%7D%20%5C%20m%2Fs)
Answer:
31.42383 m/s
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
= Coefficient of kinetic friction = 0.48
s = Displacement = 0.935 m
= Mass of bean bag = 0.354 kg
= Mass of empty crate = 3.77 kg
= Speed of the bean bag
= Speed of the crate
Acceleration
![a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g](https://tex.z-dn.net/?f=a%3D-%5Cfrac%7Bf%7D%7Bm%7D%5C%5C%5CRightarrow%20a%3D-%5Cfrac%7B%5Cmu%20mg%7D%7Bm%7D%5C%5C%5CRightarrow%20a%3D-%5Cmu%20g)
![a=--9.81\times 0.48=4.7088\ m/s^2](https://tex.z-dn.net/?f=a%3D--9.81%5Ctimes%200.48%3D4.7088%5C%20m%2Fs%5E2)
From equation of motion
![v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as%5C%5C%5CRightarrow%20v%3D%5Csqrt%7B2as%2Bu%5E2%7D%5C%5C%5CRightarrow%20v%3D%5Csqrt%7B2%5Ctimes%204.7088%5Ctimes%200.935%2B0%5E2%7D%5C%5C%5CRightarrow%20v%3D2.96739%5C%20m%2Fs)
In this system the momentum is conserved
![m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s](https://tex.z-dn.net/?f=m_1v_1%3D%28m_1%2Bm_2%29v_2%5C%5C%5CRightarrow%20v_1%3D%5Cfrac%7B%28m_1%2Bm_2%29v_2%7D%7Bm_1%7D%5C%5C%5CRightarrow%20u%3D%5Cfrac%7B%280.354%2B3.77%29%5Ctimes%202.69739%7D%7B0.354%7D%5C%5C%5CRightarrow%20u%3D31.42383%5C%20m%2Fs)
The speed of the bean bag is 31.42383 m/s
Answer:
![d=5.376km](https://tex.z-dn.net/?f=d%3D5.376km)
Explanation:
Since <em>light is so fast</em> we can assume no time passes between the lightning strikes and we observe it. We want to know then how far away did the strike occur for the sound to take 16s to reach our ears. Since the definition of velocity tells us that
, we can write ![d=vt=(336m/s)(16s)=5376m=5.376km](https://tex.z-dn.net/?f=d%3Dvt%3D%28336m%2Fs%29%2816s%29%3D5376m%3D5.376km)