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Artist 52 [7]
3 years ago
15

A baseball bat balances 74.9 cm from one end. If an 0.560-kg glove is attached to that end, the balance point moves 25.3 cm towa

rd the glove. Find the mass of the bat.
Physics
1 answer:
timurjin [86]3 years ago
3 0

Answer:

The mass of the bat is 1.09 kg.

Explanation:

Given that,

The balance point of the glove, x = 74.9 cm

Mass of the glove, m = 0.56 kg

Center of mass of the baseball bat, C = 25.3 cm

Let M is the mass of the bat. The center of mass is given by the formula as :

C=\dfrac{MX+mx}{M+m}

X is 0 as it is at a end

C=\dfrac{mx}{M+m}

25.3=\dfrac{0.56\times 74.9}{M+0.56}

M = 1.09 kg

So, the mass of the bat is 1.09 kg. Hence, this is the required solution.

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Answer:

Height, h = 16.67 m

Explanation:

We have,

Mass of a squirrel is 0.765 kg.

He jumps off the tree and hits the ground with 125 joules of energy.

It is required to find the height up on the tree the squirrel was when it jumped.

The energy possessed by the squirrel is called its gravitational potential energy. It can be given by :

E=mgh

h is height up on the tree the squirrel was when it jumped

h=\dfrac{E}{mg}\\\\h=\dfrac{125\ J}{0.765\ kg\times 9.8\ m/s^2}\\\\h=16.67\ m

So, the squirrel will go to a height of 16.67 m.

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When supernova explodes because of the collapsing cores, the core itself will become a neutron star or a white dwarf without undergoing any kind of explosive transformation. The white dwarf stars can also become supernova if they orbit another star in a binary system and steal material from their companion. In addition, the end state of a star, whether it will explode as a supernova become a black hole.

 

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How do you explain this arrangement of fossils? What might have happened?
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Answer:

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A uniform ladder of length 10.8 m is leaning against a vertical frictionless wall. The weight of the ladder is 323 N, and it mak
love history [14]

Answer:

0.3625

Explanation:

From the given information:

Consider the equilibrium conditions;

On the ladder, net torque= 0

Thus,

\tau_{net} = 0; and

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However, by rearrangement;

fL \s in \theta =m_L g \dfrac{L}{2} cos \theta + mg (7.46\ m) cos \theta  \\ \\  \mu(m_L + m) gL \ sin \theta = (323  \ N) ( 10.8 \ meters) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos  \  66.46^0

\mu= \dfrac{ (323  \ N) ( 10.8 \ m) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos  \  66.46^0}{\Big [(323 \ N)+(734 \ N) \Big] (10.8 \ m)}

\mathbf{\mu= 0.3625 }

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Answer:

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