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Simora [160]
3 years ago
9

A 70 N force pulls a box straight up 5 m above the ground. How much work is done on the box?

Chemistry
1 answer:
Bingel [31]3 years ago
5 0

D. 350 J

Explanation:

Given parameters:

Force of pull = 70N

Distance = 5m

Unknown:

Workdone on the box = ?

Solution:

Workdone is defined as the force applied on a body to move it through a distance.

   Work done = Force x distance moved

   Work done =  5 x 70 = 350J

Learn more:

Work done brainly.com/question/9100769

#learnwithBrainly

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Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product o
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Answer:

Follows are the solution:

Explanation:

A + B = C

Its response decreases over time as well as consumption of a reactants.  

r = -kAB

during response A convert into 2x while B convert into x to form 3x of C

let's  y = C

y = 3x

Still not converted sum of reaction  

for A: 100 - 2x

for B: 50 - x

Shift of x over time  

\frac{dx}{dt} = \frac{-k(100 - 2x)}{(50 - x)}

Integration of x as regards t  

\frac{1}{[(100 - 2x)(50 - x)]} dx = -k dt\\\\\frac{1}{2[(50 - x)(50 - x)]} dx = -k dt\\\\\ integral\  \frac{1}{2[(50 - x)^2]} dx =\ integral [-k ] \ dt\\\\\frac{-1}{[100-2x]} = -kt + D \\\\

D is the constant of integration

initial conditions: t = 0, x = 0

\frac{-1}{[100-2x]} = -kt + D   \\\\\frac{ -1}{[100]} = 0 + D\\\\D= \frac{-1}{100}\\\\

hence we get:

\frac{-1}{[100-2x]}= -kt -\frac{1}{100}\\\\or \\\\ \frac{1}{(100-2x)} = kt + \frac{1}{100}

after t = 7 minutes , C = 10 \ g = 3x

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Insert the above value x into \frac{1}{(100-2x)} equation = kt + \frac{1}{100} to get k.  

\to  \frac{1}{(100-2\times \frac{10}{3})}  = k \times (7) + \frac{1}{100} \\\\ \to  \frac{1}{(100- 2 \times 3.33)} =   \frac{700k + 1}{100} \\\\ \to  \frac{1}{(100-6.66)} = \frac{700k + 1}{100}\\\\ \to \frac{1}{93.34} = \frac{700k + 1}{100} \\\\

\to 100 =  93.34(700k + 1) \\\\ \to 100 =  65,338k + 700 \\\\ \to   65,338k =  -600 \\\\ \to  k =  \frac{-600}{ 65,338} \\\\ \to k= - 0.0091

therefore plugging in the equation the above value of k  

\to \frac{1}{(100-2x)} = kt +\frac{1}{100} \\\\\to \frac{1}{(100-2x)} = -0.0091t + \frac{1}{100}\\\\\to \frac{1}{(100-2x)} =  \frac{1 -0.91t}{100}\\\\\to \frac{1}{2(50-x)} =  \frac{1 -0.91t}{100}\\\\\to \frac{1}{(50-x)} =  \frac{1 -0.91t}{50}\\\\\to 50= (1-0.91t)(50-x)\\\\\to 50 = 50-45.5t-x-0.91tx\\\\\to x+0.91xt= -45.5t\\\\\to x(1+0.91t)= -45.5t\\\\\to x=\frac{-45.5t}{1+0.91t}

Let y = C

, calculate C:

y = 3x

y =3 \times \frac{-45.5t}{1+0.91t}

amount of C formed in 28 mins

x = \frac{-45.5t}{1+0.91t} , plug t = 28

\to x = \frac{-1274}{1+25.48} \\\\\to x = \frac{-1274}{26.48} \\\\\to x= -48.26

therefore amount of C formed in 28 minutes is = 3x = 144.78 grams

C: y =3 \times \frac{-45.5t}{1+0.91t}

y= 136.5 =137

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