It’s doesn’t change meaning it’s 0
Answer is 6 tires.
This is a projectile question.
First make sure units are consistent - express speed in m/s.
20 km/h = 20000m / 3600 s = 5.56 m/s
Assume the takeoff point of the ramp is at ground level (height, h, = 0m). We need to determine how long Joe is in the air, and use that time to calculate the horizontal distance he traveled.
Joe is traveling 5.56 m/s on a ramp angled at 20 degrees. There are vertical and horizontal components to his speed:
Vertical speed = 5.56sin20 = 1.90 m/s
Horizontal speed = 5.56cos20 = 5.22 m/s
An easy way to proceed is to calculate the time it takes for Joe’s vertical speed to reach 0m/s - this represents the time when Joe is at his maximum height and is therefore halfway through the trip. Double whatever time this is to find the total time of the trip. Remember he is decelerating due to gravity:
Time to peak:
a = Δv / Δt
-9.8 = -1.9 / Δt
Δt = 0.19s
Total trip time:
0.19 x 2 = 0.38s
Now that we have the total tome Joe is in the air, we can find the horizontal distance he traveled:
v = d / t
5.22 = d / 0.38
d = 1.98m
Now divide this total distance by the length of an individual tire to find the number of tires he will clear:
1.98 / 0.3 = 6.6 tires
Therefore he can jump 6 tires safely (he will land in the middle of the 7th tire).
Lots of steps I know but just try to think of the situation and keep track of the vertical and horizontal things!
Answer:
The energy dissipated as the puck slides over the rough patch is 1.355 J
Explanation:
Given;
mass of the hockey puck, m = 0.159 kg
initial speed of the puck, u = 4.75 m/s
final speed of the puck, v = 2.35 m/s
The energy dissipated as the puck slides over the rough patch is given by;
ΔE = ¹/₂m(v² - u²)
ΔE = ¹/₂ x 0.159 (2.35² - 4.75²)
ΔE = -1.355 J
the lost energy is 1.355 J
Therefore, the energy dissipated as the puck slides over the rough patch is 1.355 J
Answer:
5/2π
Explanation:
According to quizlet the answer is 5/2π
