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
Answer:
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Answer:
The turbine is rotated and rotates the generator to produce electricity.
Explanation:
Within a turbine enters the superheated steam which is at high pressure and high temperature, this steam is previously formed in the boiler when the steam enters the turbine hits each one of the blades of the turbine making it rotate at a given speed, the turbine shaft is coupled to the shaft of an electric generator and thus generates electricity.
It is also important to say that when the steam comes out of the turbine comes out at low pressure, this way the internal operating process is carried out within the turbine.
Answer:
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Answer:
The mass of the ice block is equal to 70.15 kg
Explanation:
The data for this exercise are as follows:
F=90 N
insignificant friction force
x=13 m
t=4.5 s
m=?
applying the equation of rectilinear motion we have:
x = xo + vot + at^2/2
where xo = initial distance =0
vo=initial velocity = 0
a is the acceleration
therefore the equation is:
x = at^2/2
Clearing a:
a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2
we use Newton's second law to calculate the mass of the ice block:
F=ma
m=F/a = 90/1.283=70.15 kg
