Answer:
The average speed can be calculated as the quotient between the distance travelled and the time needed to travel that distance.
To go to the school, he travels 2.4 km in 0.6 hours, then here the average speed is:
s = (2.4km)/(0.6 hours) = 4 km/h
To return to his home, he travels 2.4km again, this time in only 0.4 hours, then here the average speed is:
s' = (2.4 km)/(0.4 hours) = 6 km/h.
Now, if we want the total average speed (of going and returning) we have that the total distance traveled is two times the distance between his home and school, and the total time is 0.6 hours plus 0.4 hours, then the average speed is:
S = (2*2.4 km)/(0.6 hours + 0.4 hours)
S = (4.8km)/(1 h) = 4.8 km/h
Answer:
I don’t understand Espanol
Explanation:
sorry
Answer:
we got time and velocity over time.
so the distance is again the area underneath the graph
for a triangle with known base and height it's
4*10 / 2
distance traveled is 20
deceleration occurs when velocity decreases. that happens from t=2 till t=4
in 2 time-units we loose 10 units of velocity, so we decelerate by 5 units per 1 time
a (from t=2 to t=4) = -5v/t
You can look at groups in the same group (the columns), since they tend to have similar properties. For example, the alkali metals in group one react aggressively with water and form white compounds.
624 = 6(7x-1)
104=7x-1
105=7x
x= 15cm
