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Dafna11 [192]
4 years ago
5

Identify the careers that match the descriptions. Operates machines that cut through rocks underground so natural resources can

be harvested: Studies and manages natural resources to protect the environment: Manages forests for conservation, human enjoyment, and harvesting: Inspects, measures, and classifies logs based on quality: Operates machinery to cut down trees and move logs:
Engineering
2 answers:
Naily [24]4 years ago
7 0

Answer:

1: Mine-Cutting Machine Operator

2: Conservation Scientist

3: Forester

4: Log Grader

5: Logging Equipment Operator

Explanation:

I literally got this wrong just so i could give yall the answers lol

attashe74 [19]4 years ago
4 0

Answer:

1. Operates machines that cut through rocks underground so natural resources can be harvested: Mine-cutting and channeling operator

2. Studies and manages natural resources to protect the environment: Conservation Scientist

3. Manages forests for conservation, human enjoyment, and harvesting: Forster

4. Inspects, measures, and classifies logs based on quality: Log grader or scaler

5. Operates machinery to cut down trees and move logs: Logging equipment Operator

Explanation:

Conservation Scientist – studies and manages natural resources to protect the environment and support human uses of natural resources

Forester – manages forests for conservation, human enjoyment, and tree harvesting

Mine Cutting and Channeling Machine Operator – operates machinery in underground mines that bring natural resources up to the surface for human use

Logging Equipment Operator – operates logging machinery, such as tractors or bulldozers

Log Grader or Scaler – inspects, measures, and classifies logs based on their quality

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A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls
Iteru [2.4K]

Answer:

% increase = 26.32%

Explanation:

From conservation of mass, we can say that;

Mass flow rate at inlet = mass flow rate at exit.

Thus;

m'1 = m'2

Formula for mass flow rate is;

m' = ρV'

Where V' is volumetric flow rate = Av

Thus;

m' = ρAv

Where;

ρ is density

A is area

v is velocity

Therefore from m'1 = m'2, we can say that;

ρ1•A1•v1 = ρ2•A2•v2

Since the duct has a constant diameter, then A1 = A2

Thus, we now have;

ρ1•v1 = ρ2•v2

Making v2 the subject, we have;

v2 = ρ1•v1/ρ2

Now, since we want to find the percent increase in the velocity of the air as it flows through the dryer,we would use;

% increase = ((v2 - v1)/v1) × 100%

We have v2 = ρ1•v1/ρ2

Thus;

% increase = ((ρ1•v1/ρ2) - v1)/v1) × 100%

Factorizing v1 out, we have;

% increase = ((ρ1/ρ2) - 1)/1) × 100%

We are given;

ρ1 = 1.2 kg/m³

ρ2 = 0.95 kg/m³

Thus;

% increase = ((1.2/0.95) - 1)/1) × 100%

% increase = 26.32%

8 0
3 years ago
Name two common fuel gases that can be used for oxyfuel cutting
zlopas [31]
Hi

Acetylene and propane

I hope this help you!
8 0
1 year ago
Hellpppppppppp will give brainliest just help!!!!!!
aliya0001 [1]
Turbooooooooooi


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3 0
3 years ago
A unidirectional E-Glass fiber-epoxy composite material contains 61% by volume E-Glass fibers stressed under isostrain condition
zalisa [80]

Answer:

The total load carried by the fiber will be "98%".

Explanation:

The given values are:

V_{f}=0.61

V_{m}=1-V_{f}

     =1-0.61

     =0.39

E_{f}=10 \ Mpa

\sigma_{f}=0.35 \ Mpa

E_{m}=0.45 \ Mpa , \sigma_{m}=9\times 10^{-3} \ Mpa

As we know,

⇒  E_{e}=fE_{f}+mE_{m}

On putting the estimated values, we get

⇒       =0.61\times 10+0.39\times 0.95

⇒       =6.27 \ Mpa

Now,

⇒  \sigma_{c}=f\sigma_{f}+m\sigma_{m}

On putting the estimated values, we get

⇒       =0.61\times 0.35+0.39\times 0.009

⇒       =0.217 \ Mpa

Therefore,

The load carried by fiber,

=\frac{f\sigma_{f}}{\sigma_{c}}

=\frac{0.35\times 0.61}{0.217}

=0.98 i.e., 98%

4 0
4 years ago
Propane is to be compressed from 0.4 MPa and 360 K to 4 MPa using a two-stage compressor. An interstage cooler returns the tempe
kow [346]

Answer:

a. 81 kj/kg

b. 420.625K

c.  101.24kj/kg

Explanation:

\frac{t2}{t1} =[\frac{p2}{p1} ]^{\frac{y-1}{y} }

t1 = 360

p1 = 0.4mpa

p2 = 1.20

y = 1.13

substitute these values into the equation

\frac{t2}{360} =[\frac{1.20}{0.4} ]^{\frac{1.13-1}{1.13} }

\frac{t2}{360} =[\frac{1.2}{0.4} ]^{0.1150}\\\frac{t2}{360} =1.1347

when we cross multiply

t2 = 360 * 1.1347

= 408.5

a. the work required in the firs compressor

w=c(t2-t1)

c=1.67x10³

t1 = 360

t2 = 408.5

w = 1670(408.5-360)

= 1670*48.5

= 80995 J

= 81KJ/kg

b. n=\frac{t2-t1}{t'2-t1}

n = 80%

t2 = 408.5

t1 = 360

0.80 = 408.5-360 ÷ t'2-360

0.80 =\frac{48.5}{t'2-360}

cross multiply to get the value of t'2

0.80(t'2-360) = 48.5

0.80t'2 - 288 = 48.5

0.8t'2 = 48.5+288

0.8t'2 = 336.5

t'2 = 336.5/0.8

= 420.625

this is the temperature at the exit of the first compressor

c. cooling requirement

w = c(t2-t1)

= 1.67x10³(420.625-360)

= 1670*60.625

= 101243.75

= 101.24kj/kg

8 0
3 years ago
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