There is a missing part in the question. Found the complete text on internet:
"<span>What is the largest size vehicle (kg) it can lift if the diameter of the output line is 28.0 cm? "
Solution
The diameter of the piston is 28.0 cm, this means its radius is 14.0 cm (half the diameter), so the area of the piston is
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The maximum pressure of the lift is
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Therefore the maximum force the piston can lift is
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And the size (the mass) of the vehicle is
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<span>Carnot cycle efficiency = work done/heat supplied = (Th - Tc)/Th
where, Th is temperature of hot reservoir and Tc is temperature of cold reservoir.
we have given the values as Heat supplied = 1.3 MJ or 1300 KJ, Th = 427 degree C and Tc = 90 degree C.
converting degree Celsius to kelvin temperatures, Th = 427 + 273 = 700 K
Tc = 90 +273 = 363
solving equations, (700 - 363)/700 = work done / 1300
work done = 625.86 KJ i.e. 0.626 MJ work is done .</span>
Answer:
40 s
Explanation:
After 10 seconds, the first skater would have a 8m/s * 10s = 80 m head start
Let t be the number of seconds after the second skater starts will the second skater overtake the first skater
The distance traveled by the first skater after t seconds is
Similarly the distance traveled by the 2nd skater after t seconds is
Since the 2nd skater catches up to the 1st one after 80 m behind, the distance traveled by the 2nd one must be 80m greater than the distance of the 1st skater
We can substitute 
Answer:
40 J
Explanation:
From the question given above, the following data were obtained:
Force (F) = 10 N
Distance (s) = 4 m
Workdone (Wd) =?
Work done is simply defined as the product of force and distance moved in the direction of the force. Mathematically, we can express the Workdone as:
Workdone = force × distance
Wd = F × s
With the above formula, we can obtain the workdone as follow:
Force (F) = 10 N
Distance (s) = 4 m
Workdone (Wd) =?
Wd = F × s
Wd = 10 × 4
Wd = 40 J
Thus, 40 J of work was done.
Answer:W = 1.23×10^-6BTU
Explanation: Work = Surface tension × (A1 - A2)
W= Surface tension × 3.142 ×(D1^2 - D2^2)
Where A1= Initial surface area
A2= final surface area
Given:
D1=0.5 inches , D2= 3 inches
D1= 0.5 × (1ft/12inches)
D1= 0.0417 ft
D2= 3 ×(1ft/12inches)
D2= 0.25ft
Surface tension = 0.005lb ft^-1
W = [(0.25)^2 - (0.0417)^2]
W = 954 ×10^6lbf ft × ( 1BTU/778lbf ft)
W = 1.23×10^-6BTU
