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3 years ago

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A hoodlum throws a stone vertically downward with an initial speed of 12.0 m/s from the roof of a building,30.0 m above the grou

nd. (a) how long does it take the stone to reach the ground

1 answer:

bogdanovich [222]3 years ago

8 0

S=ut +1/2at2
30=12t+1/2*10*t2
5t2+12t-30=0

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the density of gold is 19.3 g/cm3. suppose a certain gold wedding ring deplaced 0.55mL of liquid when dropped in a glass of spar

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Answer:

19.3

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now we have

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weight of object = (19.3)(0.55)g

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specific gravity = (19.3×0.55)/(1×0.55)

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4. A student wants to determine what type of cereal his classmates like best. He buys 3 boxes of his most favorite puffed rice c

Ann [662]

D
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Lyrx [107]

Answer:

a) the maximum shear stress τ $_{max}$ the bar is 16T $_{max}$ /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴

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Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

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we know that;

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where d is the diameter

so

τ $_{max}$ = 16T $_{max}$ /πd³

Therefore, the maximum shear stress τ $_{max}$ the bar is 16T $_{max}$ /πd³

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let theta ( $\theta$ ) be the angle of twist

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now from the second image;

lets length dx which is at distance of "x" from "B"

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T(x) = tx

Elemental angle twist = d $\theta$ = T(x)dx / G $I_{p}$

so

d $\theta$ = tx.dx / G(πd⁴/32)

d $\theta$ = 32tx.dx / πGd⁴

so total angle of twist $\theta$ will be;

$\theta$ = $\int\limits^L_0 \, d\theta$

$\theta$ = $\int\limits^L_0 \,$ 32tx.dx / πGd⁴

$\theta$ = 32t / πGd⁴ $\int\limits^L_0 \, xdx$

$\theta$ = 32t / πGd⁴ [ L²/2]

$\theta$ = 16tL² / πGd⁴

Therefore, the angle of twist between the ends of the bar is 16tL² / πGd⁴

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Explanation:

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