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4 years ago

I NEED HELP ASAP!!!!!!!!!!!!

Physics

1 answer:

OLga [1]4 years ago

4 0

Answer:

Since S (displacement) = v * t

you can calculate the area under the graph

S = 1/2 * 3 * 15 - 3 * 13 = 22.5 - 39 = -16.5 m

In the triangular area the object travels at an average speed of 7.5 m/s

for 3 sec = 22.5 m

In the rectangular area the object travels at an average speed of -3 m/s

for 13 sec = -39 m.

So the net distance traveled is 22.5 - 39 = -16.5 m

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Answer:

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3 years ago

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A cannon of mass 6.43 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 7

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Answer:

The velocity of the shell when the cannon is unbolted is 500.14 m/s

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mass of shell, m₂ = 73.8-kg

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K.E = 9336032.1 J

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0 = m₁u₁ - m₂u₂

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v₂ is the final velocity of shell

$v_1 = \frac{m_2v_2}{m_1}$

Apply the principle of conservation kinetic energy

$K = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_1(\frac{m_2v_2}{m_1})^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_2v_2^2(\frac{m_2}{m_1}) + \frac{1}{2}m_2v_2^2 \\\\K = \frac{1}{2}m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\2K = m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\v_2^2 = \frac{2K}{M_2(\frac{m_2}{m_1} + 1)} \\\\v_2^2 = \frac{2*9336032.1}{73.8(\frac{73.8}{6430} + 1)}\\\\$

$v_2^2 = 250138.173\\\\v_2 = \sqrt{250138.173} \\\\v_2 = 500.14 \ m/s$

Therefore, the velocity of the shell when the cannon is unbolted is 500.14 m/s

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3 years ago

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Answer:

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Explanation:

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