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4 years ago

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a 2kg mass of copper falls from a height of 3m to an insulated floor. what is the maximum possible temperature increase of the c

opper?

1 answer:

Rainbow [258]4 years ago

6 0

The maximum possible temperature increase of the copper is $0.076^{\circ}C$

Explanation:

The energy gained by the mass of copper is equal to its change in gravitational potential energy, which is:

$E=mg \Delta h$

where

m = 2 kg is the mass of copper

$g=9.8 m/s^2$ is the acceleration of gravity

$\Delta h= 3m$ is the change in height of the copper

Substituting,

$E=(2)(9.8)(3)=58.8 J$

The energy gained by the copper is related to its change in temperature by

$E=mC_s \Delta T$

where

m = 2 kg is the mass

$C_s = 385 J/kg^{\circ}$ is the copper specific heat capacity

$\Delta T$ is the change in temperature

And solving for $\Delta T$ ,

$\Delta T = \frac{E}{mC_s}=\frac{58.8}{(2)(385)}=0.076^{\circ}C$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

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A NASA explorer spacecraft with a mass of 1,000 kg takes off in a positive direction from a stationary asteroid. If the velocity

Georgia [21]

Answer: 10000 kg

Explanation:

The momentum $p$ is given by the following equation:

$p=m.V$ (1)

Where:

$m$ is the mass of the object

$V$ is the velocity of the object

Now, in this case and according the conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}u_{1}+m_{2}u_{2}$ (2)

Where:

$m_{1}=1000kg$ is the mass of the spacecraft

$m_{2}$ is the mass of the asteroid

$v_{1}=0$ is the initial velocity of the spacecraft

$v_{2}=0$ is the initial velocity of the asteroid (because we are told the asteroid is stationary, as the spacracft is on the sateroid it remains stationary as well)

$u_{1}=250m/s$ is the final velocity of the spacecraft

$u_{2}=-25m/s$ is the final velocity of the asteroid

Rewritting (2):

$0=m_{1}u_{1}+m_{2}u_{2}$ (3)

$0=(1000kg)(250m/s)+m_{2}(-25m/s)$ (4)

Finding $m_{2}$ :

$m_{2}=10000kg$ This is the mass of the asteroid

3 0

3 years ago

Read 2 more answers

An engine extracts 441.3kJ of heat from the burning of fuel each cycle, but rejects 259.8 kJ of heat (exhaust, friction,etc) dur

Lilit [14]

Answer:

The thermal efficiency is 0.4113.

Explanation:

We know that the thermal efficiency is the ratio of work done by the engine over the heat taken

$\eta = \frac{W_{made}}{Q_{in}}$

Now, how much work the engine do in a cycle?

We know that the work done in a cycle must be equal to the heat taken minus the heat rejected

$W{made} = Q_{in} - Q_{rejected}$

So, the thermal efficiency will be:

$\eta = \frac{Q_{in} - Q_{rejected}}{Q_{in}}$

$\eta = \frac{Q_{in}}{Q_{in}} - \frac{Q_{rejected}}{Q_{in}}$

$\eta = 1 - \frac{Q_{rejected}}{Q_{in}}$

Putting the values of the problem

$\eta = 1 - \frac{259.8 kJ }{441.3kJ}$

$\eta = 0.4113$

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4 years ago

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3 years ago

(Please help, will give thanks + brainliest answer)

Bond [772]

Answer:

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Explanation:

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3 years ago