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Marat540 [252]
3 years ago
9

Your cousin has looked after you your whole life and you always looked up to him. Today he told you he sells drugs; " That's whe

re I get all the money for clothes and stuff. " He asked, Do you want to go into business with me you will make great money?" What do you do/say to him? Use the decision-making process. *
Physics
1 answer:
tatuchka [14]3 years ago
3 0

Answer:

Honestly, you should have a firm resolve not to go into drug peddling.

Politely tell him you are not interested in the business, that you are most concerned about your future and the career path you have chosen for yourself already.

He will definitely feel bad, but you can go on to advise him.

You can also advise him to stop and highlight the dangers of doing drugs.

Explanation:

There are a whole lot of disadvantages involved in the drug business,

in fact, the disadvantages out weights the advantages in the long run.

say if one is caught and sent to a life long sentence, or if one is caught up with the effect of the use of hard drugs and the health implication, to the extent that one looses his/her sanity, you see it won't make any sense anymore doing drugs in the long run. so friend you can advise you cousin to stop and thinker of good businesses that cold fetch him cool money, without the law chasing him, with that he can be a contributor to economic growth by paying his taxes too

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Answer:

(A) 570 rad

(B) 10 s

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The equations of motion for circular motions are used.

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(A)

At <em>t</em> = 2.00 s, the angular displacement, <em>θ</em>, is given by

\theta = \omega_0t+\frac{1}{2}\alpha t^2 = (30\times 2) + \frac{1}{2}\times35\times2^2=60+70 = 130\text{ rad}

After this time, it decelerates through an angular displacement of 440 rad.

Total angular displacement = 130 + 440 rad = 570 rad

(B)

At the time the circuit breaker tips, the angular velocity is given by

\omega = \omega_0+\alpha  t = 30.0+(35.0\times 2) = 30.0+70.0 =100.0\ \text{rad/s}

This becomes the initial angular velocity for the decelerating motion. Because it stops, the final angular velocity is 0 rad/s. The time for this part of the motion is calculated thus:

\theta_2 = \left(\dfrac{\omega_i+\omega_f}{2}\right)t

Here, \theta_2=440 (the angular displacement during deceleration)

The subscripts, <em>i</em> and <em>f</em>, on <em>ω</em> denote the initial and final angular velocities during deceleration.

\omega_i = 100

\omega_f = 0

t = \dfrac{2\theta_2}{\omega_i} = \dfrac{2\times400}{100} = 8\ \text{s}

This is the time for deceleration. The deceleration began at <em>t</em> = 2 s.

Hence, the wheel stops at <em>t</em> = 2 + 8 = 10 s.

(C)

The deceleration is given by

\alpha_R = \dfrac{\omega_f-\omega_i}{t} = \dfrac{0-100}{8} = -12.5\text{ rad/s}^2

The negative sign appears because it is a deceleration.

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