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aivan3 [116]
3 years ago
9

A small wooden block with mass m1 is suspended from the lower end of a light cord that is l long. The block is initially at rest

. A bullet with mass m2 is fired at the block with a horizontal velocity v0. The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of h, the tension in the cord is T0. What was the initial speed v0 of the bullet
Physics
1 answer:
nadezda [96]3 years ago
7 0

Answer:

Explanation:

The tension of the cord is not important, what matters is the height the block has risen.

When it has risen to its maximum it will have a speed of 0, and because of that it will have a kinetic energy of 0. However it will have a higher potential energy than it had at the beginning. The difference in potential energy will be:

\Delta Ep = (m1 + m2) * g * h

The energy to rise the block with the bullet embedded into it came from the kinetic energy of the bullet.

Ec = \frac{1}{2} * m2 * v0^2

These two energies are equal because all the kinetic energy the bullet had was transformed into potential energy. Therefore:

(m1 + m2) * g * h = \frac{1}{2} * m2 * v0^2

Rearranging:

V0 = \sqrt{\frac{2*(m1 + m2) * g *}{m2}}

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The Earth's radius is r_e=6.37 \cdot 10^6 m, so the meteoroid is located at a distance of:

r=r_e+2.60 r_e =3.60 r_e =  2.29 \cdot 10^7 m

And by substituting this value into the previous formula, we can find the value of g at that altitude:

g=  \frac{GM}{r^2} =  \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(2.29 \cdot 10^7 m)^2} =0.75 m/s^2

5 0
3 years ago
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Answer:

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before the flip

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ξ  = 2 × 22 × π × (1.02/2)² × 0.000047/0.2

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An atom of sodium-23 (atomic number = 11) has a positive charge of +1. Given this information, how many
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An open pipe, 0.29 m long, vibrates in the second overtone with a frequency of 1,227 Hz. In this situation, the fundamental freq
Andru [333]

Answer:

f = 409 Hz

Explanation:

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Let f is the fundamental frequency. It is given by :

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The relation between f and f₂ can be written as :

f_2=3f\\\\f=\dfrac{f_2}{3}\\\\f=\dfrac{1227}{3}\\\\f=409\ Hz

So, the fundamental frequency of the pipe is 409 Hz.              

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Answer:

g

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