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ziro4ka [17]
3 years ago
12

A piece of metal has attained a velocity of 107.8 m/sec after fallinf for 10 seconds what is its initial velocity

Physics
1 answer:
soldi70 [24.7K]3 years ago
4 0

Answer:

7.8 m/s

Explanation:

Here object is falling with a gravitational acceleration there  for we can take acceleration = 10 m/ s² and its constant through out the motion there for we can use motion equation

V = U + at

V - Final velocity

U - Initial velocity

a - acceleration

t - time

V=U+at

107.8=U + 10×10

  = 7.8 m/s

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When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam
Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s

b) the change in the combined kinetic energy of the two-car system during this collision

\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))

substitute the value in the equation above

=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J

Hence, the change in combine kinetic energy is -2534.78J

8 0
3 years ago
A 3 kg penguin is pushed by his penguin friends who give him an initial speed vo at the top of a 30 m hill. The penguin is hopin
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Answer:

This question can be answered by using conversation of energy.

K_1 + U_1 = K_2 + U_2

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Explanation:

Note that we take K_2 = 0 because we are looking for the minimum initial speed for the penguin to reach the top of the second hill. Any other speed more than this will already be enough for him.

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The displacement of the bus is 6.25 miles.
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