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GenaCL600 [577]

3 years ago

7

An object is initially moving at an unknown velocity. It accelerates at a rate of 2.0 m/s2 to a new velocity of 45 m/s in 10s. W

hat was the initial velocity?

1 answer:

katrin2010 [14]3 years ago

6 0

Answer:

Initial velocity=25m/s

Explanation:

acceleration(a)=2m/s^2

Final velocity(v)=45m/s

Time(t)=10seconds

Initial velocity(u)=?

V=u+axt

45=u+2x10

45=u+20

u=45-20

u=25m/s

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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 = $M_1\ =\ 0.900\ kg$
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mass of the larger disk = $M_2\ =\ 1.6\ kg$
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mass of the hanging block = m = 1.60 kg

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part (a)

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

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

$mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)$

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

From eqn (1) and (2), we get,

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part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

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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- Gas: in gases, molecules are totally free to move, so gases take the shape (and also the volume) of the container.

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