All categories

3 years ago

What is the horizontal distance from Point A to toe of slope?

Engineering

1 answer:

Nezavi [6.7K]3 years ago

4 0

Answer:

2.467 units

Explanation:

$Slope=\frac {\triangle y}{\triangle x}$

Therefore, slope being change in vertical to change in horizontal, we need to find the change in vertical

Change in vertical, $\triangle y=98.2-94.5=3.7$

$\frac {1 1/2}{1}=\frac {3.7}{\triangle x}$

$\triangle x=\frac {3.7\times 1}{1 1/2}\approx 2.467 units$

You might be interested in

Question 64 (1 point)

Xelga [282]

Answer: c fine sand aggregate, portland cement,fine sand

Explanation:

7 0

3 years ago

Fluorescent troffers are a type of _ lighting fixture

creativ13 [48]

The answer would be letter A

8 0

3 years ago

What kind of energy transformation happens when a boy uses energy from a sandwich to run a race

Semmy [17]

A boy eat a energy of a sandwich to run a race because when they eat a sandwich it helps them to help it mid workout and real nutritions of NYC and bring extra fuel and eating the right thing
I hope this help

4 0

3 years ago

Read 2 more answers

A heat pump cycle is used to maintain the interior of a building at 15°C. At steady state, the heat pump receives energy by heat

Hoochie [10]

Answer:

a) Ql=33120000 kJ

b) COP = 5.6

c) COPreversible= 29.3

Explanation:

a) of the attached figure we have:

HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:

W=Qh-Ql

Ql=Qh-W

where W=2000 kWh

Qh=120000 kJ/h

$Q_{l}=14days(\frac{24 h}{1 day})(\frac{120000 kJ}{1 h})-2000 kWh(\frac{3600 s}{1 h})=33120000 kJ$

b) The coefficient of performance is:

$COP=\frac{Q_{h} }{W}=\frac{120000 kJ/h*14(\frac{24 h}{1 day}) }{2000 kWh(\frac{3600 s}{1 h}) } = 5.6$

c) The coefficient of performance of a reversible heat pump is:

$COP_{reversible}=\frac{T_{h} }{T_{h}-T_{l} }$

Th=20+273=293 K

Tl=10+273=283K

Replacing:

$COP_{reversible}=\frac{293}{293-283}=29.3$

4 0

3 years ago

Given below are the measured streamflows in cfs from a storm of 6-hour duration on a stream having a drainage area of 185 mi^2.

sertanlavr [38]

Answer:

33.56 ft^3/sec.in

Explanation:

Duration = 6 hours

drainage area = 185 mi^2

constant baseflow = 550 cfs

<u>Derive the unit hydrograph using the inverse procedure </u>

first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below

Vdrh = sum of drh * duration

= 29700 * 6 hours ( 216000 secs )

= 641,520,000 ft^3.

next step : Calculate the volume of runoff in equivalent depth

Vdrh / Area = 641,520,000 / 185 mi^2

= 1.49 in

Finally derive the unit hydrograph

Unit of hydrograph = drh / volume of runoff in equivalent depth

= 50 ft^3 / 1.49 in = 33.56 ft^3/sec.in

5 0

3 years ago