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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
Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° counterclockwise from the x ‑axis,
Ivanshal [37]

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the  the second force(F₂)

F₂x= −19.0 N

F₂y=   16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }

F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }

F_{n} = 30.11N

Direction of the net force (β)

\beta =tan^{-1} (\frac{29.13}{7.62} )

β=75.3°

Magnitude and direction of the net force

Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis

In the attached graph we can observe the magnitude and direction of the net force

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3 years ago
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Galina-37 [17]

Answer:

100nm

Explanation:

Work=Force×distance

=20×5

=100nm

6 0
2 years ago
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