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3 years ago

15

If there were no air resistance, how long would it take a free-falling skydiver to fall from a plane at 3500 m to an altitude of

400 m , where she will pull her ripcord?

1 answer:

xxMikexx [17]3 years ago

7 0

By equation of motion we have S = ut + 0.5 a $t^{2}$

Where u = Initail velocity, S= dispalcement, t = time taken, a = acceleration

Here S = 3500-400 = 3100m, u = 0m/s, g = 9.8 m/ $s^{2}$

So, 3100 = 0+0.5*9.8* $t^{2}$

$t^{2}$ = 3100/4.9

t = $\sqrt{3100/4.9}$

t = 25.15 seconds

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Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec

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Answer:

(a) $F_{net}$ = 4.19 x $10^{-5}$ N, and its direction is towards $m_{2}$ .

(b) It must be placed inside a hollow shell.

Explanation:

Let, $m_{1}$ = 165 kg, $m_{2}$ = 465 kg, $m_{3}$ = 60 kg, and the distance between $m_{1}$ and $m_{2}$ is 0.340 m.

(a) Since $m_{3}$ is placed midway between $m_{1}$ and $m_{2}$ , then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = $\frac{Gm_{1}m_{2} }{r^{2} }$

Where all variables have their usual meaning.

Then,

a. $F_{net}$ = $F_{23}$ - $F_{13}$

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$F_{23}$ = $\frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }$

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∴ $F_{net}$ = = 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

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The net force exerted by the two masses on the 60 kg object is 4.19 x $10^{-5}$ N.

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= 4.19 x $10^{-5}$ N

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3 years ago

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

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Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$

mass of the larger disk = $M_2\ =\ 1.6\ kg$
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part (a)

A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

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Net torque on the smaller disk,

$\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)$

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Let ' $a_2$ ' be the acceleration of the block in the second case,

From the above expression,

$\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2$

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