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1 year ago

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A ball is thrown up into the air with 100 j of kinetic energy, which is transformed to gravitational potential energy at the top

of its trajectory. The ball encounters some air resistance. What is its kinetic energy when it returns to its original height?.

1 answer:

xenn [34]1 year ago

8 0

The kinetic energy when it returns to its original height is 100 J

The ball is thrown up with a Kinetic Energy K. E. = 0.5×m×v² = 100 J

Therefore the final height is given by

u² = v² -2·g·s

Where:

u = final velocity = 0

v = initial velocity

s = final height

Therefore v² = 2·g·s = 19.62·s

P.E = Potential Energy = m·g·s

Since v² = 2·g·s

Substituting the value of v² in the kinetic energy formula, we obtain

K. E. = 0.5×m×2·g·s = m·g·s = P.E. = 100 J

When the ball returns to the original height, we have

v² = u² + 2·g·s

Since u = 0 = initial velocity in this case we have

v² = 2·g·s and the Kinetic energy = 0.5·m·v²

Since m and s are the same then 0.5·m·v² = 100 J.

Learn more about kinetic energy at:

brainly.com/question/25959744

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God bless... hope this help to clear your doubt.

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3 years ago

A 1.0 kg ball and a 2.0 kg ball are connected by a 1.1-m-long rigid, massless rod. The rod is rotating cw about its center of ma

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The torque that will bring the balls to halt, t = 0.34 Nm

What is rotating?

In physics and mechanics, torque is the rotational equivalent of linear force. It is also referred to as the moment, moment of force, rotational force, or turning effect, depending on the field of study. It represents the capability of a force to produce a change in the rotational motion of the body.

Center of Mass of Rod = (1 * 0 + 2.0 * 1.1) /(1.0 + 2.0)

Center of Mass of Rod = 0.73 m

ω = 21 rpm

ω = 21 * (2*pi)/60 rad/s

ω = 2.2 rad/s

Time to bring Halt = 5.2 s

Angular De-acceleration = 2.2/5.2 rad/s^2

α = 0.423 rad/s^2

We know,

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Substituting Values -

torque = 0.8067 Kg*m^2 * 0.423 rad/s^2

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1 year ago

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3 years ago

docker41 [41]

<h3>Answer :</h3>

Let the final temperature be "T".

For the piece of copper :

mass, $\sf{m_c=40\ g.}$

specific heat capacity, $\sf{c_c=0.4\ J\,g^{-1}\,K^{-1}.}$

initial temperature, $\sf{T_c=200^{\circ}C.}$

Then the heat of copper :

$\sf{\dashrightarrow Q_c=m_cc_c\,\Delta\!T_c}$

$\sf{\dashrightarrow Q_c =16(T-200)\ J}$

For copper calorimeter :

mass, $\sf{m_{cc} =60\ g.}$

specific heat capacity, $\sf{c_{cc} =0.4\ J\,g^{-1}\,K^{-1}.}$

initial temperature, $\sf{T_{cc} =25^{\circ}C.}$

Then the heat of copper calorimeter :

$\sf{\dashrightarrow Q_{cc} =m_{cc}c_{cc}\,\Delta\!T_{cc}}$

$\sf{\dashrightarrow Q_{cc} =24(T-25)\ J}$

For water :

mass, $\sf{m_w=50\ g. }$

specific heat capacity, $\sf{c_w= 4.2\ J\,g^{-1}\,K^{-1}.}$

initial temperature, $\sf{T_w=25^{\circ}C.}$

Then heat of water :

$\sf{\dashrightarrow Q_w=m_wc_w\,\Delta\!T_w}$

$\sf{\dashrightarrow Q_w=210(T-25)\ J}$

By energy conservation, the sum of all these energies should be zero as there were no heat energy change before the process, i.e.,

$\sf{\dashrightarrow Q_c+Q_{cc}+Q_w=0}$

$\sf{\dashrightarrow16(T-200)+24(T-25)+210(T-25)=0}$

$\sf{\dashrightarrow 250T- 9050=0}$

$\sf{\dashrightarrow T=36.2^{\circ}C}$

$\large \underline{\underline{\boxed{\sf T=36.2^{\circ}C}}}$

<u>____________________________</u>

[Note: in case of considering temperature difference it's not required to convert the temperatures from $\sf{^{\circ}C}$ to K or K to $\sf{^{\circ}C}$ .]

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3 years ago