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2 years ago

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Neglecting the effect of air resistance a stone dropped off a 175-m high building lands on the ground in: A)3s B)4s C)6s D)18s E

)36s

1 answer:

Bad White [126]2 years ago

5 0

Answer:

c) 6s

Explanation: take downwards as positive.

let vf = the velocity of the stone when it hits the ground, vi = 0 be the starting velocity of the stone. r0 = 175m be the height of the building.

(vf)^2 = (vi)^2 + 2g(r-r0)

(vf)^2 = 2g(r0)

vf = \sqrt{2g(r0)}

= \sqrt{2×9.8×(175)}

= 58.6 m/s

then:

vf = vi + g×t

vf = g×t

t = vf/g

= (58.6)/(9.8)

= 6s

Therefore, it will take the stone 6 seconds to reach the gound.

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A circular pipe of 25-mm outside diameter is placed in an airstream at 25C and 1-atm pressure. The air moves in cross flow over

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Answer:

f_D = =3.24 N/m

Explanation:

data given

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$\rho = 1.048 kg/m3$

k = 0.0288 W/m.K

WE KNOW THAT

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= 1.941 *10^4

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$C_D = \frac{f_D}{A_f\frac{\rho v^2}{2}}$

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$f_D = C_D A_f*\frac{\rho v^2}{2}$

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Drag coffecient for smooth circular cylinder is 1.1

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$f_D = 1.1*0.025 *\frac{1.048*15^2}{2}$

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Marcus can drive his boat 24 miles down the river in 2 hours but takes 3 hours to return upstream. Find the rate of the boat in

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Answer:

speed of boat as

$v_b = 10 mph$

river speed is given as

$v_r = 2 mph$

Explanation:

When boat is moving down stream then in that case net resultant speed of the boat is given as

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$v_1 = v_r + v_b$

Now when boat moves upstream then in that case the net speed of the boat is opposite to the speed of the river

so here we have

$v_2 = v_b - v_r$

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$v_1 = \frac{24}{2} = 12 mph$

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Answer:

The value of charge q₃ is 40.46 μC.

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Given that.

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Using formula of net force

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