<span>To begin, the mouse walks from 5 to 12 cm, for a displacement of 7 cm. Next, it walks 8 cm in the opposite direction, for a total displacement of (7 + [-8]) or (-1) cm. This leaves the mouse on 4 cm, and then it walks from there to the 7cm location, for a displacement of 7-4 or +3 cm. Adding 3cm to -1cm gives a final displacement of +2cm.</span>
Answer:
c > √(2ab)
Explanation:
In this exercise we are asked to find the condition for c in such a way that the results have been real
The given equation is
½ a t² - c t + b = 0
we can see that this is a quadratic equation whose solution is
t = [c ±√(c² - 4 (½ a) b)] / 2
for the results to be real, the square root must be real, so the radicand must be greater than zero
c² -2a b > 0
c > √(2ab)
Answer:
d= 794.4 cmExplanation:
Given that
Speed ,V= 286 km/h
V=79.44 m/s
Given that time ,t= 100 ms
t= 0.1 s
We know that ( if acceleration is zero)
Distance = Speed x time
d= V t
Now by putting the values in the above equation
d = 79.44 x 0.1 m
d= 7.944 m
We know that 1 m = 100 cm
d= 794.4 cm
I don't think anyone knows the answer
