The first car travels at 60km/h and skid at 30m away from the starting point while another car is also traveling at 180km/h. Now, we need to solve for the skidding distance.
We assigned variables such as:
V1=60km/h
V2=180km/h
Skid1=30m
Skid2=?
We solve this by ratio and proportion method such as shown below:
V1/V2=skid1/skid2
60/180=30/skid2
skid2=(30*180)/60
skid2=90meters
Th answer is 90 meters.
Power = work/time = (Force times distance)/time
= (30N *10.0m)/5.00s = 300/5 = 60 Watts
On the dog's return trip (between <em>t</em> = 10 and <em>t</em> = 12.5 seconds), the slope of the position function is steeper than during the first 5 seconds, which means the dog ran home faster. The only option that captures this is D.
You can check to make sure that the dog indeed runs twice as fast on the return trip. The slope of the position function during the first 5 seconds is
(change in position) / (change in time) = (5 - 0) / (5 - 0) = 5/5 = 1
while during the return trip, it is
(0 - 5) / (12.5 - 10) = -5/2.5 = -2
Ignoring the sign (which only indicates the direction in which the dog was running), we see that the dog's speed on the return trip was indeed twice as high as during the first 5 s.
Answer:
B
Magnified images will not be created.
Explanation:
I did it and this was the correct answer
