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BigorU [14]
3 years ago
8

Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig

ht of tower.​
Physics
1 answer:
HACTEHA [7]3 years ago
4 0

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

y=-\frac{1}{2} gt^2+v_0t+y_0

With the given values in the question the equation has one unknown v₀:

v_0=\frac{y-y_0}{t}+\frac{1}{2}gt

Solving for t=1:

1) v_0=y-y_0+\frac{g}{2}

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) E=\frac{1}{2}m{v_0}^2+mgy_0

If h is the height of the tower, the energy on top of the tower:

3) E=mgh

Combining equation 2 and 3 and solving for h:

4) h=\frac{{v_0}^2}{2g}+y_0

Combining equation 1 and 4:

h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0

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Answer:

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Explanation:

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6 0
3 years ago
The trough of the sine curve used to represent a sound wave corresponds to
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Answer:

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Have a nice day!

4 0
3 years ago
A 0.350kg bead slides on a curved fritionless wire,
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Answer:

h2 = 0.092m

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Since the collision is elastic, we now that velocity of bead 1 after the collision is given by:

V_{B'}=V_B*\frac{m1-m2}{m1+m2} = \sqrt{2gh}* \frac{m1-m2}{m1+m2}=-1.34316m/s

Now, by doing another balance of energy from the instant after the collision, to the point where bead 1 stops, we get the distance it rises:

m1*g*h2-\frac{m1*V_{B'}^2}{2}=0 Solving for h2:

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70\omega^2=3.90 \frac{m}{s^2}  \\  \\ \omega= \sqrt{ \frac{3.9}{70} }

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