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1 year ago

How much work does it take to lift 345 boxes to a height of 6.00 m of each box has a mass of 7.89 kg

1 answer:

3 0

Given the value of the mass of each boxes, the work done in lifting the boxes to the given height is 1.6 × 10⁵J.

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

Work done W = ?

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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Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

$y = \frac{1}{2} at^2+v_0t+y_0$

Where,

a = acceleration

t = time

$v_0$ = Initial velocity

$y_0$ = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

$v = \frac{x}{t}$

x = Displacement

t = time

With the data we have we can find the time as well

$v = \frac{x}{t}$

$t = \frac{x}{v}$

$t = \frac{18.6}{34}$

$t = 0.547s$

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

$y = \frac{1}{2} at^2+v_0t+y_0$

$y = \frac{1}{2} gt^2+0+0$

$y = \frac{1}{2} gt^2$

$y = \frac{1}{2} 9.8*0.547^2$

$y = 1.46m$

Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.

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2 years ago

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First of all, we know that one mole is equal to the atomic number of an element.

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And we need to find 5 moles.

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I really hope this gave you and ideas and helped you in some way:)

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2 years ago

You toss a bowling ball straight up into the air with a speed of 2.1 m/s. How long does it take the bowling ball to reach its hi

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Time taken by the bowling ball to reach its highest point= 0.214 s

initial velocity= Vi=2.1 m/s

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using the kinematic equation Vf= Vi + at

0= 2.1 + (-9.8)t

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melamori03 [73]

Answer:

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Explanation:

In hyperopia ,people face difficulties to see close up object , but can see object easily which are at a distance.

The main reason of hyperopia is our eyeball.When our eyeball become too short , then light focus behind the retina. Sowe will face problem to see near object but we can see distance object easily. Hyperopia is the opposite of nearsightedness. Hyperopia can be corrected by using contact lenses.

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