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ryzh [129]
3 years ago
9

Melissa uses bags of mulch that weigh 40 pounds each. She needs to carry 10 bags to her truck. She usually carries one bag at a

time. Would she do less work if she carried two bags at a time?
No; she would have to use half the force over twice the distance.

Yes; she makes fewer trips, decreasing the distance.

No; she would have to use twice the force over half the distance.

Yes; she decreases both force and distance.
Chemistry
2 answers:
KatRina [158]3 years ago
7 0

Answer: Option (a) is the correct answer.

Explanation:

Work done is related to force as force times displacement.

Mathematically,           W = F × d

where       W = work

                 F = force

                 d = displacement

Therefore, when Melissa carried two bags at a time then she is applying twice the force she was applying earlier. Hence, she have to apply half the force twice the distance.

Thus, we can conclude that no; she would have to use half the force over twice the distance.

umka21 [38]3 years ago
6 0
NO:she would have to use twice the force over half the distance
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<em>Answer:</em>

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What is the density of a substance that weighs 100g and has a volume of 68cm3?
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1. If there are 100 navy beans, 27 pinto beans and 173 blackeyed peas in a container, what is the percent abundance in the conta
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The answers to the two questions are:

1. The percent abundance in the container which has <u>100 navy</u>, <u>27 pinto</u>, and <u>173 black-eyed peas beans</u> is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed <em>peas </em>beans, respectively.

2. The <em>weighted </em>average score for the scores of 85, 75, 96 obtained from the evaluations of exams (20%), labs (75%), and homework (96%) is 84.1.      

1. The percent abundance by type of bean is given by:

\% = \frac{n}{n_{t}} \times 100   (1)

Where:

n: is the number of each type of beans

n_{t}: is the <em>total number</em> of <em>beans </em>

The <em>total number </em>of <em>beans</em> can be calculated by adding the number of all the types of beans:

n_{t} = n_{n} + n_{p} + n_{b}   (2)

Where:

n_{n}: is the number of navy beans = 100

n_{p}: is the number of pinto beans = 27

n_{b}: is the number of black-eyed <em>peas </em>beans = 173  

Hence, the total number of beans is (eq 2):

n_{t} = 100 + 27 + 173 = 300  

Now, the <em>percent abundance</em> by type of bean is (eq 1):

  • Navy beans

\%_{n} = \frac{100}{300} \times 100 = 33.3 \%

  • Pinto beans

\%_{p} = \frac{27}{300} \times 100 = 9.0 \%

  • Black-eyed peas beans

\%_{b} = \frac{173}{300} \times 100 = 57.7 \%

Hence, the percent abundance by type of bean is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed peas beans, respectively.

2. The average score (S) can be calculated as follows:

S = e*\%_{e} + l*\%_{l} + h*\%_{h}   (3)

Where:

e: is the score for exams = 85

l: is the score for lab reports = 75

h: is the score for homework = 96

%_{e}\%_{e}: is the <em>percent</em> for<em> exams</em> = 70.0%

\%_{l}: is the <em>percent </em>for <em>lab reports</em> = 20.0%

\%_{h}: is the <em>percent </em>for <em>homework </em>= 10.0%

Then, the <u>average score</u> is:

S = 85*0.70 + 75*0.20 + 96*0.10 = 84.1

We can see that if the score for each evaluation is 100, after multiplying every evaluation for its respective percent, the final average score would be 100.  

Therefore, the <em>weighted </em>average score will be 84.1.

Find more about percents here:

  • brainly.com/question/255442?referrer=searchResults
  • brainly.com/question/22444616?referrer=searchResults

I hope it helps you!

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At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?
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Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

\Delta H\ for\ diamond = 1.897 kJ/mol

\Delta H\ for\ graphite = 0 kJ/mol

\Delta S\ for\ diamond = 2.38 J/(K mol)

\Delta S\ for\ graphite = 5.73 J/(K mol)

(a) We need to calculate the value of \Delta G for diamond

Using formula of Gibbs free energy change

\Delta G=\Delta H-T\Delta S

Put the value into the formula

\Delta G= (1897-0)-298\times(2.38-5.73)

\Delta G=2895.3

\Delta G=2.895\ kJ

The Gibbs free energy  change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

\Delta S=V\times\Delta P

\Delta S=\dfrac{m}{\rho}\times\Delta P

Put the value into the formula

\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130

\Delta S=34.59\ J/mole

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

\beta= \Delta G_{g}+\Delta G+\Delta G_{d}

\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}

\beta=\Delta P(V_{d}-V_{g})+\Delta G

Put the value into the formula

0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3

0=-0.0194\Delta P+2895.3

\Delta P=\dfrac{2895.3}{0.0194}

\Delta P=14924\ atm

(d). Here, C_{p}=0

So, The value of \Delta H and \Delta S at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

\Delta G=\Delta H-T\Delta S

Put the value into the formula

\Delta G=(1897-0)-900\times(2.38-5.73)

\Delta G=4912\ J

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

7 0
3 years ago
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