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poizon [28]
3 years ago
14

A leaky 10-kg bucket is lifted from the ground to a height of 10 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

ly the bucket contains 30 kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 10-m level. Find the work done. (Use 9.8 m/s2 for g.) Show how to approximate the required work by a Riemann sum. (Let x be the height in meters above the ground. Enter xi* as xi.)

Physics
1 answer:
sladkih [1.3K]3 years ago
8 0

Answer:

The work done shall be 14715 Joules

Explanation:

The work done by a force 'F' in a displacement 'dy' is given by

W=m(y)g\times dy

At any position 'y' the weight shall be sum of weft of water and weight of string

\therefore m(y)=m_{water}(y)+m_{string}(y)\\\\m(y)=30(1-\frac{y}{10})+0.9y

Thus applying values we get

W=\int m(y)g\times dy\\\\W=\int_{0}^{10}(30(1-\frac{y}{10}+0.9y)\times g)dy\\\\Solving\\W=294.3\int_{0}^{10}(1-\frac{y}{10}+0.9y)dy\\\\W=14715J

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Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.
DiKsa [7]

Answer:0.061

Explanation:

Given

T_C=300 k

Temperature of soup T_H=340 K

heat capacity of soup c_v=33 J/K

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any  instant

efficiency is given by

\eta =\frac{dW}{Q}=1-\frac{T_C}{T}

dW=Q(1-\frac{T_C}{T})

dW=c_v(1-\frac{T_C}{T})dT

integrating From T_H to T_C

\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT

W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT

W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}

W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]

Now heat lost by soup is given by

Q=c_v(T_C-T_H)

Fraction of the total heat that is lost by the soup can be turned is given by

=\frac{W}{Q}

=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}

=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}

=\frac{300-340-300\ln (\frac{300}{340})}{300-340}

=\frac{-40+37.548}{-40}

=0.061

4 0
3 years ago
Why is cheese only strechy when its hot not when its cold (i spelt stretchy wrong but i dont want to change it cuz that would ta
IgorC [24]
That is a really good question, cheese is stretchy when it is hot is because when you heat it up, it liquefies which makes it stretch. it doesn't stretch when it is cold because it is a solid and solids usually do not stretch.
8 0
3 years ago
An 75-kg hiker climbs to the summit of Mount Mitchell in western North Carolina. During one 2.0-h period, the climber's vertical
Ostrovityanka [42]

Answer:

The change in gravitational potential energy of the climber-Earth system is  \Delta  PE  = 396900 \ J

Explanation:

From the question we are told that

    The mass of the hiker is  m = 75 \ kg

    The time  taken is  T  =  2 \ hr =  2 *  3600 =  7200 \ s

    The  vertical elevation after time  T is  H = 540 \ m

   

The  change  in gravitational potential is  mathematically represented as

         \Delta  PE  =  mgH

here g is the acceleration due to gravity with value  g =  9.8 \ m/s^2  

     substituting values  

        \Delta  PE  =  75  *  9.8  *  540

       \Delta  PE  = 396900 \ J

3 0
3 years ago
The cooling effect inside a refrigerator is produced by
nlexa [21]
Actually, when the refrigeration liquid is vaporized, the cooling effect is produced
6 0
3 years ago
In 1977 off the coast of Australia, the fastest speed by a vessel on the water
fenix001 [56]

Answer: 154.08 m/s

Explanation:

Average acceleration a_{ave} is the variation of velocity  \Delta V over a specified period of time  \Delta t:

a_{ave}=\frac{\Delta V}{\Delta t}}

Where:

a_{ave}=1.80 m/s^{2}

\Delta V=V_{f}-V_{o} being V_{o}=0 the initial velocity and V_{f} the final velocity

\Delta t=85.6 s

Then:

a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}}

Since V_{o}=0:

a_{ave}=\frac{V_{f}}{\Delta t}}

Finding V_{f}:

V_{f}=a_{ave} \Delta t

V_{f}=(1.80 m/s^{2})(85.6 s)

Finally:

V_{f}=154.08 m/s

8 0
3 years ago
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