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Gala2k [10]
4 years ago
6

g Suppose Howard is pulling a bucket of bricks up along the side of a building with a rope. The bricks have a mass of 20 kg and

are not touching the side of the building. With what tension force must Howard achieve to (briefly) keep an acceleration rate of 2.0 m/s2 on the bucket of bricks?
Physics
1 answer:
cupoosta [38]4 years ago
4 0

Answer:

= 236N

Explanation:

tension T = mg + ma

Given that,

m = 20kg

g = 9.8 m/s²

a = 2.0 m/s²

T = m(g + a)

T = 20( 9.8 + 2.0)

  = 20(11.8)

  = 236N

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Answer:

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From the free fall movement we have the following formulas: Vf^{2} = Vo^{2} - 2gh and h=Vo*t - \frac{g*t^{2} }{2}, First we need to find the height to time iqual to 3.04 s using the formula: h=Vo*t - \frac{g*t^{2} }{2} and remember that golf ball was released from the rest (Vo= 0 (m/s)) so h= (0 (m/s))*(3.04 (s)) - \frac{9.8 (m/s^2)*(3.04 (s))^{2} }{2}, we get: h = -45.28 (m) with the height that we have got, now the velocity of the ball is calculate using Vf^{2} = Vo^{2} - 2gh solving for Vf, we get: Vf = \sqrt{Vo^{2}-2*g*h } replacing the values given Vf = \sqrt{(0 m/s)^{2}-2*(9.8 m/s^2)*(-45.28 m) }, so we get: Vf = 29.79 (m/s).

5 0
4 years ago
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PLS HELP
Ahat [919]
<h3>B. True</h3>

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3 years ago
If a material has a high thermal conductivity, is it a<br> conductor or an insulator?
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cestrela7 [59]

Answer:

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Explanation:

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So if we start with such formula with a given mass, radius, and tangential velocity, and then we move to a situation where everything stays the same except for the radius which doubles, then the new centripetal force (F'_c) will be given by: F'_c=m\,* \frac{v^2}{2r}

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