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2 years ago

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A firecracker breaks up into two pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.0 m/

s. The second piece has a mass of 300 g and flies off along the y-axis with a speed of 45.0 m/s. What is the total momentum of the two pieces?

1 answer:

larisa [96]2 years ago

4 0

Answer:

Total momentum, p = 21.24 kg-m/s

Explanation:

Given that,

Mass of first piece, $m_1=200\ g= 0.2\ kg$

Mass of the second piece, $m_2=300\ g= 0.3\ kg$

Speed of the first piece, $v_1=82\ m/s$ (along x axis)

Speed of the second piece, $v_2=45\ m/s$ (along y axis)

To find,

The total momentum of the two pieces.

Solve,

The total momentum of two pieces is equal to the sum of momentum along x axis and along y axis.

$p_x=m_1v_1$

$p_x=0.2\ kg\times 82\ m/s$

$p_x=16.4\ kg-m/s$

$p_y=m_2v_2$

$p_y=0.3\ kg\times 45\ m/s$

$p_y=13.5\ kg-m/s$

The net momentum is given by :

$p=\sqrt{p_x^2+p_y^2}$

$p=\sqrt{16.4^2+13.5^2}$

p = 21.24 kg-m/s

Therefore, the total momentum of the two pieces is 21.24 kg-m/s.

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Since, the normal stress in concrete is induced as a result of temperature rise; we have the expression :

$\sigma_c =\frac{P}{A_c}$

$\sigma_c = \frac{6482.98}{96.39}$

$\sigma_c = 67.26 psi$

Thus, the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

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$\sigma_s = \frac{-P}{A_s}$

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