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raketka [301]
3 years ago
13

Two students, Student 1 and Student 2, did a hands-on mining exercise with cookies to compare the total cost of mining in two di

fferent land areas using different equipment. The table below shows the costs of land, equipment, mining, and reclamation for the two students.

Physics
1 answer:
grandymaker [24]3 years ago
3 0
Assume cost of equipment for student 2 is m.

We are given that:
Total mining cost for student 1 and student 2 is $40

For student 1:
We are given that:
Cost of land = $5
Cost of equipment = $2
Cost of mining and reclamation = $2 per minute
Time taken to mine = 5 minutes
Therefore,
Total mining cost for student 1= land + equipment
                                                  + cost of mining (time needed)
Total mining cost for student 1 = 2 + 5 + 2(5) = 7 + 10 = $17

This means that:
Total mining cost for student 2 = 40 - 17 = $23

For student 2:
We are given that:
Cost of land = $4
Cost of equipment = m
Cost of mining and reclamation = $2 per minute
Time taken to mine = 6 minutes
Total mining cost for student 2= $23
Therefore,
Total mining cost for student 2= land + equipment
                                                  + cost of mining (time needed)
23 = 4 + m + 2(6)
23 = 4 + m + 12
23 = 16 + m
m = 23 - 16
m = $7

Based on the above calculations, the correct choice would be:
c. $7
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3 years ago
A cord is attached to the box and run through a pulley directly above the box, so that the cord is vertical. The free end of the
Harman [31]

Answer:

The answer is given here would be a simplified equation, seeing as there are some missing variables in the question.

<u>F1 = T- 46, 674.656 gm/s² </u>

Explanation:

<em>Note: Once we have the mass of the second object and/or acceleration of the cord, we can solve for the force of the ground acting on the box.</em>

To calculate the force caused by gravity on the basic pulley system we use the following equation:

F2 = M2 x g; where g= gravitational acceleration (a constant equal to 9.8 m/s²). The mass M2 = 10.5 lb = 4762.72g

∴ F2 = 4762.72g x 9.8 m/s²

= 46, 674.656 gm/s² or 46, 674.656 N

But since this F2 is acting in a downlowrd direction, it would be negative.

Tension of the cord, T = Mass, x × acceleration. ( x is in the pulley diagram)

⇒ F1 = T - F2

<u>F1 = T- 46, 674.656 gm/s² </u>

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3 years ago
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Hope I helped!

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3 years ago
Read 2 more answers
a man is trying to pull a box a distance of 3 m with a force of 20 N that makes a 35º with the horizontal.
Ivanshal [37]

Answer:

34.4Joules

Explanation:

Complete question

a man is trying to pull a box a distance of 3 m with a force of 20 N that makes a 35º with the horizontal. Find the workdone

Work done = Fdsin theta

Force F = 20N

distance d = 3m

theta = 35 degrees

Substitute

Workdone = 20(3)sin 35

Workdone = 60sin35

Workdone = 34.4Joules

Hence the workdone by the man is 34.4Joules

5 0
2 years ago
A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is
Lana71 [14]

Answer:

a

   m  = 0.169 \ kg

b

  |v_{max} |=  0.5653 \ m/s

Explanation:

From the question we are told that

    The  spring constant is  k =  14 \ N/m

     The  maximum extension of the spring is  A =  6.0 \ cm  =  0.06 \ m

     The number of oscillation is  n  =  30

      The  time taken is  t  =  20 \ s

Generally the the angular speed of this oscillations is mathematically represented as

           w = \frac{2 \pi}{T}

where T is the period which is mathematically represented as

     T  =  \frac{t}{n}

substituting values

     T  =  \frac{20}{30 }

     T  = 0.667 \ s

Thus  

       w = \frac{2 * 3.142 }{ 0.667}

       w =  9.421 \ rad/s

this angular speed can also be represented mathematically as

       w =  \sqrt{\frac{k}{m} }

=>   m  =\frac{k }{w^2}

substituting values

      m  =\frac{ 15 }{(9.421)^2}

      m  = 0.169 \ kg

In SHM (simple harmonic motion )the equation for velocity is  mathematically represented as

        v =  - Awsin (wt)

The  velocity is maximum when  wt = \(90^o) \ or \ 1.5708\ rad

     v_{max} = -  A* w

=>   |v_{max} |=  A* w

=>    |v_{max} |=   0.06 * 9.421

=>   |v_{max} |=  0.5653 \ m/s

5 0
3 years ago
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