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joja [24]
3 years ago
10

A load of mass 5kg is raised through a height of 2m. calculate the work done against (g=10mls)​

Physics
1 answer:
Illusion [34]3 years ago
3 0

The work done against gravity is 100 J

Explanation:

The work done against gravity in order to lift an object is equal to the change in gravitational potential energy of the object:

W=mg\Delta h

where

m is the mass of the object

g is the acceleration of gravity

\Delta h is the change in height of the object

For the object in this problem, we have:

m = 5 kg

g=10 m/s^2

\Delta h = 2 m

Substituting into the equation,

W=(5)(10)(2)=100 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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Potential energy becomes kinetic energy when:
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Suppose you have two solid bars, both with square cross-sections of 1 cm2. They are both 24.6 cm long, but one is made of copper
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Explanation:

Expression to calculate thermal resistance for iron (R_{I}) is as follows.

             R_{I} = \frac{L_{I}}{k_{I} \times A_{I}}  

where,   L_{I} = length of the iron bar

             k_{I} = thermal conductivity of iron

             A_{I} = Area of cross-section for the iron bar

Thermal resistance for copper (R_{c}) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]

where,  L_{c} = length of copper bar

             k_{c} = thermal conductivity of copper

            A_{c} = Area of cross-section for the copper bar

Now, expression for the transfer of heat per unit cell is as follows.

           Q = \frac{(100^{o} - 0^{o}}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}

 Putting the given values into the above formula as follows.

       Q = \frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}

  = \frac{(100^{o} - 0^{o})}{21 \times 10^{-2} m[\frac{1}{73 \times 10^{-4}m^{2}} + \frac{1}{386 \times 10^{-4}m^{2}}}

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It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,

                 P = \frac{Q}{T}

Here, T is 1 second so, power conducted is equal to heat transferred.

So,           P = 2.92 watt

Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

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