1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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
You might be interested in
Most energy obtained from water is converted from _____.
Snowcat [4.5K]
Potential energy behind dams
7 0
3 years ago
Read 2 more answers
If you lift a five pound object 18 inches how many joules of energy did yo use?
kogti [31]
The work is 90 as 5 times 18

4 0
3 years ago
If two micro coulomb of charge is flowing in a circuit for 5 minutes. What is the amount of current in the circuit?
Murrr4er [49]

Answer:

6.67×10¯⁹ A

Explanation:

From the question given above, the following data were obtained:

Quantity of electricity (Q) = 2 μC

Time (t) = 5 mins

Current (I) =?

Next, we shall convert 2 μC to C. This can be obtained as follow:

1 μC = 1×10¯⁶ C

Therefore,

2 μC = 2 μC × 1×10¯⁶ C / 1 μC

2 μC = 2×10¯⁶ C

Next, we shall convert 5 mins to seconds. This can be obtained as follow:

1 min = 60 secs

Therefore,

5 min = 5 min × 60 sec / 1 min

5 mins = 300 s

Finally, we shall determine the current in the circuit. This can be obtained as follow:

Quantity of electricity (Q) = 2×10¯⁶ C

Time (t) = 300 s

Current (I) =?

Q = It

2×10¯⁶ = I × 300

Divide both side by 300

I = 2×10¯⁶ / 300

I = 6.67×10¯⁹ A

Thus, the current in the circuit is 6.67×10¯⁹ A

4 0
3 years ago
What is the density of a 150cm 3 block with a mass of 700 grams?
Vinvika [58]

Answer:

The Density of the block is 4.667g/mL

Explanation:

Given the following data;

Mass of block = 700g

Volume of block = 150cm³

Density = ?

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the equation;

Density = \frac{mass}{volume}

Substituting into the equation, we have;

Density = \frac{700}{150}

<em>Density = 4.667g/mL</em>

3 0
3 years ago
A slender rod is 80.0 cm long and has mass 0.390 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05
Keith_Richards [23]

Answer

given,

length of slender rod =80 cm = 0.8 m

mass of rod = 0.39 Kg

mass of small sphere = 0.0200 kg

mass of another sphere weld = 0.0500 Kg

calculating the moment of inertia of the system

I = \dfrac{ML^2}{12}+\dfrac{m_1L^2}{4}+\dfrac{mL^2}{4}

I = \dfrac{0.39\times 0.8^2}{12}+\dfrac{0.02\times 0.8^2}{4}+\dfrac{0.05\times 0.8^2}{4}

I =0.032\ kg.m^2

using conservation of energy

\dfrac{1}{2}I\omega^2 = (m_1-m_2)g\dfrac{L}{2}

\omega=\sqrt{\dfrac{(m_1-m_2)gL}{I}}

\omega=\sqrt{\dfrac{(0.05-0.02)\times 9.8 \times 0.8}{0.032}}

\omega=2.71 \rad/s

we know,

v = r ω

v = \dfrac{L}{2} \times 2.71

v = \dfrac{0.8}{2} \times 2.71

v = 1.084 m/s

3 0
3 years ago
Other questions:
  • Select the true statements: check all that apply check all that apply oxidizing agents can convert co into co2. a reducing agent
    15·1 answer
  • According to a rule-of-thumb. every five seconds between a lightning flash and the following thunder gives the distance to the f
    10·1 answer
  • A uniform disk turns at 5.00 rev/s around a frictionless spindle. A non-rotating rod, of the same mass as the disk and length eq
    10·1 answer
  • Which vector should be negative?
    11·1 answer
  • Light incident on a surface at an angle of 45° undergoes diffused reflection. At what angle will it reflect?
    5·2 answers
  • One event in a high school track meet is the 400 meter run. If the runners of the 4oo meter event start and end at the same loca
    12·1 answer
  • This image shows a stream of positively charged particles being directed at gold foil. The positively charged particles are call
    13·2 answers
  • Difference between precion and accuarcy
    7·1 answer
  • A real image can be obtained with:
    7·2 answers
  • It takes a truck 3.56 seconds to slow down from 112 km/h to 87.4 km/h. What is its average acceleration?
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!