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Given Data: Diameter 'd' = 30 cm = 0.3 m Lifting Weight 'W' = mg = 2000*9.81 N = 19,620 N Calculations: Area of the lift 'A' = <span>pi\over4*d^2=pi\over4*0.3^2=0.07 m^2
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Given the value of the mass of each boxes, the work done in lifting the boxes to the given height is 1.6 × 10⁵J.
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Work done</h3>
Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;
W = F × d
Where F is force applied or Weight and d is distance
Also Force = Weight = mass × acceleration due to gravity.
Since gravity is acting on the boxes as it been lift
W = Weight × height from ground level
W = mg × d
Where m is mass of the boxes, g is accelration due to gravity( g = 9.8m/s² ) and d is distance from ground level.
Given the data in the question;
- Since each box has a mass of 7.89 kg
- Mass of the 345 boxes = 345 × 7.89 kg = 2722.05kg
- Distance or height d = 6.0m
To determine the work done, we substitute our values into the expression above.
W = mg × d
W = 2722.05kg × 9.8m/s² × 6.0m
W = 160056.5kgm²/s²
W = 160056.5J
W = 1.6 × 10⁵J
Therefore, Given the value of the mass of each boxes, the work done in lifting the boxes to the given height is 1.6 × 10⁵J.
Learn more about work done here: brainly.com/question/26115962
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Answer:
The focal length of the appropriate corrective lens is 35.71 cm.
The power of the appropriate corrective lens is 0.028 D.
Explanation:
The expression for the lens formula is as follows;
Here, f is the focal length, u is the object distance and v is the image distance.
It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.
Put v= -71.4 cm and u= 24.0 cm in the above expression.
f= 35.71 cm
Therefore, the focal length of the corrective lens is 35.71 cm.
The expression for the power of the lens is as follows;
Here, p is the power of the lens.
Put f= 35.71 cm.
p=0.028 D
Therefore, the power of the corrective lens is 0.028 D.
