We want to know what does the fact that Mercury has no atmosphere tell us. Since Mercury has no atmosphere it cant reflect a lot of sunlight that is hitting its surface. Its constantly being hit by solar wind. So Mercury reflects a small percentage of the sunlight that strikes it.
Answer:
<em> B.0</em>
Explanation:
Change in momentum: This is defined as the product of mass and change in velocity of a body. or it can be defined as the product of force and time of a body. The fundamental unit of change in momentum is kg.m/s
Change in momentum = M(V-U)......................... Equation 1
where M = mass of the ball, V = final velocity of the ball, U = initial velocity of the ball.
Let: M = m kg and V = U = v m/s
Substituting these values into equation 1
Change in momentum = m(v-v)
Change in momentum = m(0)
Change in momentum = 0 kg.m/s
<em>Therefore the momentum of the ball has not changed.</em>
<em>The right option is B.0</em>
Answer:
The observed frequency by the pedestrian is 424 Hz.
Explanation:
Given;
frequency of the source, Fs = 400 Hz
speed of the car as it approaches the stationary observer, Vs = 20 m/s
Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.
The observed frequency is calculated as;
![F_s = F_o [\frac{v}{v_s + v} ] \\\\](https://tex.z-dn.net/?f=F_s%20%3D%20F_o%20%5B%5Cfrac%7Bv%7D%7Bv_s%20%2B%20v%7D%20%5D%20%5C%5C%5C%5C)
where;
F₀ is the observed frequency
v is the speed of sound in air = 340 m/s
![F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\](https://tex.z-dn.net/?f=F_s%20%3D%20F_o%20%5B%5Cfrac%7Bv%7D%7Bv_s%20%2B%20v%7D%20%5D%20%5C%5C%5C%5C400%20%3D%20F_o%20%5B%5Cfrac%7B340%7D%7B20%20%2B%20340%7D%20%5D%20%5C%5C%5C%5C400%20%3D%20F_o%20%280.9444%29%20%5C%5C%5C%5CF_o%20%3D%20%5Cfrac%7B400%7D%7B0.9444%7D%20%5C%5C%5C%5CF_o%20%3D%20423.55%20%5C%20Hz%20%5C%5C)
F₀ ≅ 424 Hz.
Therefore, the observed frequency by the pedestrian is 424 Hz.