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4 years ago

I need help with this question

Physics

1 answer:

rusak2 [61]4 years ago

8 0

Answer:

The answer is C

Explanation:

That's because all has the same vertical velocity, but C has the highest horizontal velocity.

I hope I could help.

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A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force does work in raising th

LuckyWell [14K]

Answer:

25032.47 W

Explanation:

Power is the time rate of doing work, hence,

P = Work done(non conservative) / time

Work done (non conservative) is given as:

W = total K. E. + total P. E.

Total K. E. = 0.5mv²- 0.5mu²

Where v (final velocity) = 7.0m/s, u (initial velocity) = 0m/s

Total P. E. = mgh(f) - mgh(i)

Where h(f) (final height) = 7.2m, h(i) (initial height) = 0 m

=> W = 0.5mv² - mgh(f)

P = [0.5mv² - mgh(f)] / t

P = [(0.5*790*7²) - (790*9.8*7.2)] / 3

P = (19355 + 55742.4) / 3 = 75097.4/3

P = 25032.47 W

6 0

3 years ago

1) In which of these examples is the greatest movement occurring? A) A B) B C) C D) D

Hatshy [7]

Answer:d

Explanation:

8 0

3 years ago

Slow-twitch muscles are needed for what?

Tresset [83]

Slow-twitch muscles help enable long-endurance feats such as distance running, while fast-twitch muscles fatigue faster but are used in powerful bursts of movements like sprinting. Hope that this can help!!!

8 0

3 years ago

Read 2 more answers

PLEASE HELP!!!!

Goryan [66]

Answer:

The braking distance would be about nine times as long (assuming that acceleration during braking stays the same.)

Explanation:

Let $u$ denote the initial velocity of the vehicle ( $20\; \text{mph}$ or $60\; \text{mph}$ ) and let $v$ denote the velocity of the vehicle after braking ( $0\; \text{mph}$ ). Let $x$ denote the braking distance.

Assume that the acceleration during braking are both constantly $a$ in both scenarios. The SUVAT equations would apply. In particular:

$\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a}\end{aligned}$ .

Since $v = 0$ (the vehicle has completely stopped), the equation becomes $x = (-u^{2}) / (2\, a)$ .

Assuming that $a$ (braking acceleration) stays the same, the braking distance $x$ would be proportional to $u^{2}$ , the square of the initial velocity.