Missing graph. I attach it in the answer.
In a uniformly accelerated motion, the velocity at time t is given by:

where a is the acceleration and t is the time.
Given the previous equation, if we plot v(t) versus t, we find a straight line; moreover, a (the acceleration) represents the slope of the curve.
Looking at the graph, we see that when the time goes from 10 s to 20 s, the velocity increases from 4 m/s to 6 m/s. Therefore the slope of the curve is

and this corresponds to the acceleration.
So, the correct answer is <span>
0.2 m/s2.</span>
Because Mars is too far away for its gravitational pull to affect us, in addition Earths gravitational pull is greater than Mars anyways.
Option B & Option D is your correct answers.
'cause here, atom has eight electrons in it's valence shell, so it means it has stable structure which falls in group 18
Hope this helps!
Answer:
.
Explanation:
Distance Skylar traveled North is 
Then she traveled
Westward.
After which she traveled
towards the South.
The total distance traveled would be the sum of the distances.

The distance traveled by Skylar was is
.
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.