Answer:
The correct answer to the following question will be Option A (moment arm; pivot point).
Explanation:
- The moment arm seems to be the duration seen between joint as well as the force section trying to act mostly on the joint. Each joint that is already implicated in the workout seems to have a momentary arm.
- The moment arm extends this same distance from either the pivot point to just the position of that same pressure exerted.
- The pivotal point seems to be the technical indicators required to fully measure the appropriate demand trends alongside different time-frames.
The other three choices are not related to the given situation. So that option A is the appropriate choice.
Acceleration, a = (v - u)/t
where v is the final velocity, u is the initial velocity, and t is the time.
This formula on a velocity time graph represents the slope of the graph.
Answer:
Hello there Dude answer is B :D hope it helped mark me brainliest.
Answer:
When the speed of the bottle is 2 m/s, the average maximum height of the beanbag is <u>0.10</u> m.
When the speed of the bottle is 3 m/s, the average maximum height of the beanbag is<u> 0.43</u> m.
When the speed of the bottle is 4 m/s, the average maximum height of the beanbag is <u>0.87</u> m.
When the speed of the bottle is 5 m/s, the average maximum height of the beanbag is <u>1.25</u> m.
When the speed of the bottle is 6 m/s, the average maximum height of the beanbag is <u>1.86</u> m.
Sorry for not answering early on! If anyone in the future needs help, I got these answers from 2020 egenuity, though I can't post the picture for proof. Stay Safe!
Answer:
a) a= 8.33 m/s², T = 12.495 N
, b) a = 2.45 m / s²
Explanation:
a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.
We apply Newton's second law to the lower charge
fr₁ + fr₂ = ma
The equation for the force of friction is
fr = μ N
Y Axis
N - W₁ –W₂ = 0
N = W₁ + W₂
N = (m₁ + m₂) g
Since the beams are the same, it has the same mass
N = 2 m g
We replace
μ₁ 2mg + μ₂ mg = m a
a = (2μ₁ + μ₂) g
a = (2 0.30 + 0.25) 9.8
a= 8.33 m/s²
Let's look for cable tension with beam 2
T = m₂ a
T = 1500 8.33
T = 12.495 N
b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal
fr = m₂ a₂
N- w₂ = 0
N = W₂ = mg
μ₂ mg = m a₂
a = μ₂ g
a = 0.25 9.8
a = 2.45 m / s²