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tiny-mole [99]
3 years ago
13

A cardinal (Richmondena cardinalis) of mass 4.20×10−2 kg and a baseball of mass 0.146 kg have the same kinetic energy. What is t

he ratio of the cardinal's magnitude pc of momentum to the magnitude pb of the baseball's momentum?
Physics
1 answer:
bekas [8.4K]3 years ago
6 0

Explanation:

It is given that,

Mass of cardinal,m_c=4.2\times 10^{-2}\ kg

Mass of baseball, m_b=0.146\ kg

Both cardinal and baseball have same kinetic energy. We need to find the ratio of the cardinal's magnitude p_c of momentum to the magnitude p_c of the baseball's momentum.

E_c=E_b

Kinetic energy is given by, E=\dfrac{p^2}{2m}

\dfrac{p_c^2}{2m_c}=\dfrac{p_b^2}{2m_b}

(\dfrac{p_c}{p_b})^2=\dfrac{m_c}{m_b}

\dfrac{p_c}{p_b}=\sqrt{\dfrac{m_c}{m_b}}

\dfrac{p_c}{p_b}=\sqrt{\dfrac{4.2\times 10^{-2}}{0.146}}

So, \dfrac{p_c}{p_b}=\dfrac{53}{100}

So, the ratio of cardinal's magnitude of momentum to the magnitude of the baseball's momentum is 53 : 100

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Shalnov [3]

Answer is

9.773m/s^2

-----------------------------------------------------------------------------

Given,

h=8848m

The value of sea level is 9.08m/s^2. So, Let g′ be the acceleration due to the gravity on Mount Everest.

g′=g(1 − 2h/h)

=9.8(1 - 6400000/17696)

=9.8(1 − 0.00276)

9.8×0.99724

=9.773m/s^2

Thus, the acceleration due to gravity on the top of Mount Everest is =9.773m/s^2

-----------------------------------------------------------------------

hope this helps :)

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2 years ago
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tresset_1 [31]

Answer:

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You throw a ball upward with a speed of 14m/s. What is the acceleration of the ball after it leaves your hand? Ignore air resist
omeli [17]

The acceleration of the ball after leaving the hand is 9.8 m/s^2 downward

Explanation:

In order to find the acceleration of the ball during its motion, we have to study which forces are acting on it.

After the ball leaves the hand, if we neglect air resistance, there is only one force acting on the ball: the force of gravity, whose magnitude is

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According to Newton's second law, the acceleration of the ball is given by

a=\frac{\sum F}{m}

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\sum F is the net force acting on the ball

After the ball leaves the hand, the only force acting on it is the force of gravity, so we can substitute (mg) into the previous equation:

a=\frac{mg}{m}=g=9.8 m/s^2

This means that the acceleration of the ball remains 9.8 m/s^2 downward for the entire motion, after leaving the hand.

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brainly.com/question/3820012

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