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NeX [460]
3 years ago
9

You are on a Parkour course. First you climb a angled wall up 9.5 meters. They you shimmy along the edge of a 3.5 meter long wal

l. Lastly you traverse 15 meters of obstacles to reach the end of the course. The trip takes 43 seconds. What is your average speed?
0.22 m/s


1204 m/s


0.65 m/s


1.5 m/s
Physics
1 answer:
makvit [3.9K]3 years ago
4 0

Average speed = (total distance covered) / (time to cover the distance)

Total distance covered = (9.5m + 3.5m + 15m) = 28 meters

Time to cover the distance = 43 seconds

Average speed = (28 meters) / (43 seconds)

Average speed = 0.65 meters/second

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21. Calculate the acceleration of the bus from point D to E. Show your work.
Marat540 [252]

21) Acceleration from D to E: 1 m/s^2

22) The acceleration of the bus from D to E is 1 m/s^2

Explanation:

21)

The acceleration of an object is equal to the rate of change of velocity of the object. Mathematically:

a=\frac{v-u}{t}

where

u is the initial velocity

v is the final velocity

t is the time elapsed

In this problem, we want to measure the acceleration of the bus from point D to point E. We have:

- Initial velocity at point D: u = 0

- Final velocity at point E: v = 5 m/s

- Time elapsed from D to E: t = 21 - 16 = 5 s

Therefore, the acceleration between D and E is

a=\frac{5-0}{5}=1 m/s^2

22) This question is the same as 21), so the result is the same.

Learn more about acceleration:

brainly.com/question/9527152

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4 0
3 years ago
A hockey puck is sliding across a frozen pond with an initial speed of 9.3 m/s. It comes to rest after sliding a distance of 42.
kondaur [170]

Answer:

The coefficient of kinetic friction between the puck and the ice is 0.11

Explanation:

Given;

initial speed, u = 9.3 m/s

sliding distance, S = 42 m

From equation of motion we determine the acceleration;

v² = u² + 2as

0 = (9.3)² + (2x42)a

- 84a = 86.49

a = -86.49/84

|a| = 1.0296

F_k = \mu_k N = ma

where;

Fk is the frictional force

μk is the coefficient of kinetic friction

N is the normal reaction = mg

μkmg = ma

μkg = a

μk = a/g

where;

g is the gravitational constant = 9.8 m/s²

μk = a/g

μk = 1.0296/9.8

μk = 0.11

Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11

3 0
3 years ago
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