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3 years ago

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A bus accelerates at -0.5 m/s2 for 12.5 seconds . If the bus had an initial speed of 31 m/s, what is its new speed? A. 12.25m/s

B. 37.25m/s C. 24.75m/s D. 6.25m/s

1 answer:

KonstantinChe [14]3 years ago

4 0

Answer:

37.5m/s

Explanation:

a=(v-u)/t

0.5 = (v - 31) / 12.5

v - 31 = 0.5 x 12.5

v - 31 = 6.25

V = 6.25 + 31

v = 37.25m/s

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Describe an example where materials are prepared as fluids so that they can be moved easily

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A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

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Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

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Mass of the second puck = $m_2\ =\ 3\ kg$
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Part (a)

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$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

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part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

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Given,

Time of impact = t = $25\times 10^{-3}\ sec$

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9. A radioisotope has a half-life of 4.50 min and an initial decay rate of 8400 Bq. What will be

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Answer:

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t is the time,

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