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galina1969 [7]
3 years ago
6

Assignment Using Perman's equation, estimate the potential evapotranspiration for the month of August in locality with the follo

wing data
latitude 20 degrees north
Elevation 200m
• Mean monthly temperature 20 degreeso
• Mean relative humidity: 75%
• Mean sunshine hour= 9hrs
Wind at 2 m height equal 85 km per day
nature of surface cover =green grass​
Engineering
1 answer:
Korolek [52]3 years ago
3 0

Answer:

what do you need help with

Explanation:

You might be interested in
Read the first paragraph of the article.
Paladinen [302]

Answer:

i think it is A, i could not really understand your question.

Explanation:

just a question

(wheres the article?)

5 0
3 years ago
If welding is being done in the vertical position, the torch should have a travel angle of?
siniylev [52]

Answer:

Between 35°– 45°

Explanation:

In the vertical position, Point the flame in the direction of travel. Keep the flame tip at the correct height above the base metal. An angle of 35°–45° should be maintained between the torch tip and the base metal. This angle may be varied up or down to heat or cool the weld pool if it is too narrow or too wide

4 0
2 years ago
If the length of the arm were to increase, how would the required torque change? a. Required torque would increase b. Required t
monitta
B.requires torque would decrease (I think)
4 0
3 years ago
The torque required to drive a 40 mm diameter power screw having double square threads with a pitch of 6 mm is 1.0 kN-m. The fri
myrzilka [38]

Answer:

This is self locking.

Explanation:

Given that

Diameter d= 40 mm

Pitch (P)= 6 mm

Torque T= 1 KN.m

Friction(μ) = 0.1

Thread is double square.

Condition for self locking

\mu >tan\theta

tan\theta=\dfrac{L} {\pi d}

L= n x P

N= number of start

Here given that double start so n=2

L=2 x 6 = 12 mm

tan\theta=\dfrac{L} {\pi d}

tan\theta=\dfrac{12} {\pi \times 40}

tan\theta=0.095            (μ = 0.1)

\mu >tan\theta

So from we can that this is self locking.

6 0
3 years ago
3 cm of water evaporated from a 200-hectare vertical walled reservoir during 24 hours. Storm water was added to the reservoir at
ycow [4]

Answer:

water released through the bottom of the reservoir in 24 hrs is 1992 ha-cm

Explanation:

Given the data in the question;

A = 200 hectare =  2 × 10⁶ m²

water evaporated Ve = 2 × 10⁶ m² × 3 × 10⁻² = 60000 m³ { in 24 hrs }

Water added by storm in 24hrs Vi = 3 × 24 × 3600 = 259200 m³

now let water released be Vr

ΔV = V_ini - V_final = 0

Vi - Ve - Vr = 0

Vr = Vi - Ve

Vr = 259200 m³ - 60000 m³

Vr = 199200 m³ = 19920000 m² - cm

Vr = 1992 ha-cm

Therefore, water released through the bottom of the reservoir in 24 hrs is 1992 ha-cm

7 0
3 years ago
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