Answer:
d. less than 20m/s
Explanation:
To the 2nd car, the first car is travelling 10m/s east and 10m/s south. So the total velocity of the first car with respect to the 2nd car is
[tex]\sqrt{10^2 + 10^2} =10\sqrt{2}=14.14m/s
As 14.14m/s is less than 20m/s. d is the correct selection for this question.
Answer: a) 456.66 s ; b) 564.3 m
Explanation: The time spend to cover any distance a constant velocity is given by:
v= distance/time so t=distance/v
The slower student time is: t=780m/0.9 m/s= 866.66 s
For the faster students t=780 m/1,9 m/s= 410.52 s
Therefore the time difference is 866.66-410.52= 456.14 s
In order to calculate the distance that faster student should walk
to arrive 5,5 m before that slower student, we consider the follow expressions:
distance =vslower*time1
distance= vfaster*time 2
The time difference is 5.5 m that is equal to 330 s
replacing in the above expression we have
time 1= 627 s
time2 = 297 s
The distance traveled is 564,3 m
A campfire being lighted and plants converting carbon-dioxide and water into glucose and oxygen are both forms of chemical change.
Therefore, the answer is:
B. Both are examples of chemical change.
Answer:
D40 = 2.56 × D25
so number is 2.56 multiple of stopping distance @ 25 mph
Explanation:
given data
speed = 40 miles / hour
distance = D40
speed limit = 25 miles / hour
distance = D25
to find out
express number a multiple of stopping distance @ 25 mph
solution
we know here stopping distance is directly proportional to (speed)²
so here speed ratio is
initial speed = 
so initial speed = 1.6
so
stopping distance increase = (1.6)²
= (1.6)²
= 2.56
so here
D40 = 2.56 × D25
so number is 2.56 multiple of stopping distance @ 25 mph
