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2 years ago

Do you think the mining process is faster when you know in advance that the land must be restored? Explain.

Engineering

1 answer:

vfiekz [6]2 years ago

3 0

Answer: No, the mining process isn't faster when you know in advance that the land must be restored.

Explanation:

The mining process would be slower when mining companies know in advance that the land must be restored because they have to be more careful about their impact on the environment & avoid taking unnecessary damage to ruin the land which they know they must restore.

So, it's evident how this would slow down the mining process. If there were no obligations to restore land, the mining companies would just come at the mining space and run their mining process fast without big concerns for the consequences.

But the caution definitely slows them down.

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the velocity of a particle is given by v=16t^2i+4t^3j+(5t+2k) m/s, where t is in seconds. if the particle is at the origin when

pshichka [43]

Answer:

80.16 m/s^2

at t=2 s

x=42.3 m

y=16 m

z=14 m

Explanation:

solution

The x,y,z components of the velocity are donated by the i,j,k vectors.

$v_{x}=16t^{2} \\v_{y}=4t^{3}\\v_{z}=5t+2$

acceleration is a derivative of velocity with respect to time.

$a_{x}=\frac{d}{dt} v_{x}=\frac{d}{dt}[16t^{2}]=32t\\a_{y}=\frac{d}{dt} v_{y}=\frac{d}{dt}[4t^{3}]=12t^{2} \\a_{z}=\frac{d}{dt} v_{z}=\frac{d}{dt}[5t+2]=5$

evaluate acceleration at 2 seconds

$a_{x} =32*2=64m/s^{2}\\ a_{y} =12*2^{2} =48m/s^{2}\\a_{z} =5m/s^{2}$

the magnitude of the acceleration is the square root of the sum of the square of each component of the acceleration.

$=\sqrt{a_{x}^2 +a_{y}^2+a_{z} ^2 } \\=\sqrt{64^2 +48^2+5 ^2 }\\=80.16m/s^2$

position is the integral of velocity with respect to time position at a time can be found by taking by taking the definite intergral of each component.

$x=\int\limits {v_{x} } \, dx=\int\limits^2_0 {{16t^2} \, dt=42.7m\\\\y=\int\limits {v_{y} } \, dx=\int\limits^2_0 {{4t^3} \, dt=16m\\\\\\\\\\z=\int\limits {v_{z} } \, dx=\int\limits^2_0 {{5t+2} \, dt=14m\\\\$

3 0

2 years ago

If something is 50fficient, how many joules of wasted energy will there be if 750j of energy is put in?’

Sedbober [7]

Efficiency is the minimum use of energy to accomplish the task. The wasted energy will be 375 J when 750 J of energy is given.

<h3>What is wasted energy?</h3>

Wasted energy is energy that is not useful when the transformation in the system occurs.

Total energy = 750 J

The efficiency of the system = 50 %

Output work (OW) is calculated as:

Efficiency = output work ÷ input work × 100%

750 × 50 = 100 OW

OW = 375 J

Wasted energy = Total energy - output work

= 750 - 375

= 375 J

Therefore, the machine is 50 % inefficient and has wasted energy of 375 J.

Learn more about wasted energy here:

brainly.com/question/16177264

#SPJ4

8 0

1 year ago

In software engineering how do you apply design for change?

Pavlova-9 [17]

Answer:

it is reducely very iloretable chance for a software engineer to give an end to this question

7 0

2 years ago

Line(s) indicates passing is allowed if there are no oncoming cars.

anygoal [31]

Broken yellow b/c you can’t pass on a double solid yellow

5 0

3 years ago

The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with di

jeka57 [31]

Answer:

a) at T = 5800 k

band emission = 0.2261

at T = 2900 k

band emission = 0.0442

b) daylight (d) = 0.50 μm

Incandescent ( i ) = 1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

<em>Values are gotten from the table named: blackbody radiati</em>on functions

<u>a) Calculate the band emission fractions for the visible region</u>

at T = 5800 k

band emission = 0.2261

at T = 2900 k

band emission = 0.0442

attached below is a detailed solution to the problem

<u>b)calculate wavelength corresponding to the maximum spectral intensity</u>

For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

3 0

2 years ago