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____ [38]
2 years ago
11

An artillery shell of mass 30 kg has a velocity of 250 m/s vertically upward. The shell explodes into two pieces; immediately af

ter the explosion a fragment of mass 10 kg has a velocity of 120 m/s straight downward. How high above the point of the explosion does the larger fragment rise?
Physics
1 answer:
olga_2 [115]2 years ago
7 0

Answer:

9654.34 m

Explanation:

from conservation of momentum

$$\begin{aligned}30 \times 250 &=-10 \times 120+20 \times V \\20 V &=30 \times 250+10 * 120 \\V &=\frac{30 \times 250+10 \times 120}{20}=435 \mathrm{~m} / \mathrm{s}\end{aligned}$$

And from Conservation of Energy

\frac{1}{2} m v^{2}=m g h\\h=\frac{v^{2}}{2 g}\\h=\frac{(435(m/s))^{2}}{2 \times 9.8(m/s^{2} )}\\h=9654.34 (m)

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Answer:

<h3> 3.057m</h3>

Explanation:

According to law of gravitation;

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G is the universal gravitation

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d² = 6.67408 × 10-11 *3000*7000/0.0015

d² = 140.15568*10^-5/0.0015

d² = 1.4016*10^-3/0.0015

d² = 1.4016*10^-3/1.5*10^-3

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nikklg [1K]

Answer:

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Explanation:

Acceleration is defined as the change in velocity divided by the time it took to produce such change. The formula then reads:

a = \frac{change-in-velocity}{time} = \frac{Vf-Vi}{t}

Where Vf is the final velocity of the object, (in our case 80 m/s)

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and t is the time it took to change from the Vi to the Vf (in our case 0.05 seconds.

Therefore we have:

a = \frac{80 m/s - 0 m/s}{0.05 sec} = 1600 \frac{m}{s^2}

Notice that the units of acceleration in the SI system are \frac{m}{s^2} (meters divided square seconds)

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