Answer:
2500 and kg its is very easy do with method
Answer:
2.61 J
Explanation:
Since potential energy U = mgy where m = mass of object, g = acceleration due to gravity = 9.8 m/s² and y = height of object above the ground.
Now for the coffee mug, m= 0.422 kg and it is 0.63 m on a table, so it is 0.63 m above the ground. Thus, y = 0.63 m.
We compute U
U = mgy
= 0.422 kg × 9.8 m/s² × 0.63 m
= 2.605 J
≅ 2.61 J
So, the potential energy of the mug with respect to the floor is 2.61 J
Given the value of the mass of each boxes, the work done in lifting the boxes to the given height is 1.6 × 10⁵J.
<h3>
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
you must have a rock first of all then the cycle continues
Answer:
Explanation:
The horizontal distance covered by the ball in the falling is only determined by its horizontal motion - in fact, it is given by
where
is the horizontal velocity
t is the time of flight
The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.
In the first situation, the horizontal distance covered is
in the second case, the horizontal velocity is increased to
And so the new distance travelled will be
So, the distance increases linearly with the horizontal velocity.
