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3 years ago

À stone is thrown

1 answer:

3 0

A = -9.8

v = -9.8t -8

s = -4.9 t2 -8t +25

So… -5t^2 -8t + 25 =0, we’ll rearrange to 5t^2 + 8t - 25. We get two roots, one is positive and is 1.59 seconds

V = -9.8(1.59) - 8 = -23.6

So… it takes 1.59 seconds to hit the ground at -23.6 m/s.

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Indique onde em quantos metros o rapaz chegará com as seguintes condições: S0 = 0 a) v = 3 m/s e t = 2 s b) v = 2 m/s e t = 3,5

Naily [24]

Answer:

I will answer in English.

Here we will use the relation

Velocity*time = distance

So:

a) velocity = 3m/s

time = 2s

Distance = 3m/s*2s = 6m

b) velocity = 2m/s

time = 3.5s

Distance = 2m/s*3.5s = 7m

c) velocity = 10m/s

time = 0.5s

Distance = 10m/s*0.5s = 5m

d) velocity = 4m/s

time = 2.5s

Distance = 4m/s*2.5s = 9m

e) velocity = 1.5m/s

time = 5s

Distance = 1.5m/s*5s = 7.5m

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3 years ago

HELP!!!What type of image is formed by a lens if m=-0.87?

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3 years ago

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A typical laboratory centrifuge rotates at 4000 rpm. Test tubes have to be placed into a centrifuge very carefully because of th

cestrela7 [59]

Answer:

1. $a_{rad}=17545.2\frac{m}{s^{2}}$

2. $a=4429.45 \frac{m}{s^{2}}$

Explanation:

Radial acceleration is:

$a_{rad}=\frac{v^2}{r}$ (1)

With r the radius respects the axis of rotation and v the tangential velocity that is related with angular velocity (ω) by:

$v= \omega r$ (2)

By (2) on (1):

$a_{rad}= \frac{(\omega r)^2}{r}= (\omega )^2r=(418,88\frac{rad}{s})^2(0.1m)$

$a_{rad}=17545.2\frac{m}{s^{2}}$

To find the acceleration of the tube with the fall, we can use the expression:

$\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t)$ (3)

Due impulse-momentum theorem:

$\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i}$ (4)

with p the momentum and J the impulse. By (4) on (3):

$\overrightarrow{p}_{f}-\overrightarrow{p}_{i}=\overrightarrow{F}_{avg}(\varDelta t)$

And using Newton's second law (F=ma) and that (P=mv):

$mv_f-mv_i=(ma)(\varDelta t)$ (5)

Final velocity is the velocity just after the encounter with hard floor, and initial momentum us just before that moment so the first one is zero and the second one can be found sing conservation of energy:

$\frac{mv_i}{2}=mgh$