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2 years 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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3 years ago

Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long. How long is the ot

s344n2d4d5 [400]

Answer:

The length of the longer pipe is L = 2.30 m

Explanation:

Given that:

Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long.

How long is the other pipe?

From above;

The formula for the frequency of open ended pipes can be expressed as:

$f = \dfrac{nv}{2L}$

where n = 1 ( since half wavelength exist between those two pipes)

v = 343 m/s and L = 2.08 m

Thus, the shorter pipe produces a frequency of :

$f = \dfrac{1*343}{2*2.08}$

$f = \dfrac{343}{4.16}$

$f =82.45 \ Hz$

Also; we know that the beat frequency was given as 8.0 Hz

Then,

The lower frequency of the longer pipe = ( 82.45 - 8.0 )Hz

The lower frequency of the longer pipe = 74.45 Hz

Finally;

From the above equation; make Length L the subject of the formula. Then,

The length of the longer pipe is L = $\dfrac{nv}{2f}$

The length of the longer pipe is L = $\dfrac{1*343}{2*74.45}$

The length of the longer pipe is L = $\dfrac{343}{148.9}$

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3 years ago

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3 years ago

if a sine wave representing ac voltage has an effective value of 70 vac what would be the waves peak value

Morgarella [4.7K]

Answer:

99 V

Explanation:

The effective voltage of an AC current (also called rms voltage) is given by

$V_{rms} = \frac{V_0}{\sqrt{2}}$

where

Vrms is the rms voltage

V0 is the peak voltage

In this problem, we know the effective voltage:

$V_{rms} = 70 V$

Therefore, we can re-arrange the equation to find the peak voltage:

$V_0 = V_{rms} \sqrt{2} =(70 V)(\sqrt{2})=99 V$

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3 years ago