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3 years ago

How are weather and climate different

Engineering

2 answers:

fomenos3 years ago

6 0

Answer:

- weather refers to short-term changes in the atmosphere

- climate describes what the weather is like over a long period of time in one area.

Svetradugi [14.3K]3 years ago

3 0

Whereas weather refers to the short term changes in the atmosphere, climate describes what the weather is like over a long period of time in a specific area. Weather tells you what you want to wear each day. Climate tells you what types of clothes you have in your closet. Hope it helps. Pls mark branliest have a great day

You might be interested in

Write a program to input 6 numbers. After each number is input, print the biggest of the numbers entered so far.

likoan [24]

Answer:

P<u>rogram:</u>

# Enter Numbers #

number1 = int(input("Enter number: " ))

print("Largest: " + string(number1))

#for num 2 #

number2 = int(input("Enter a number: "))

if number2 > number1:

print("Largest: " + string(number2))

else:

print("Largest: " + string(num1))

#for num 3 #

number3 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3)))

#for num 4 #

number4 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4)))

#for num 5 #

number5 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4, number5)))

#for num 6 #

number6 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4, number5, number6)))

# END #

4 0

3 years ago

A tension member must be designed for a service dead load of 18 kips and a service live load of 2 kips.

ella [17]

Answer:

A) max factored load ( pv = 1.4 * 18 ) = 25.2 kips

B) max load factored load = ( Pa = 18 + 2 ) = 20 kips

Explanation:

service dead load = 18 kips

service live load = 2 kips

A) Determine the maximum factored load and controlling AISC load combination

max factored load ( pv = 1.4 * 18 ) = 25.2 kips

DL = 18 kips

LL = 2 kips

B) Determine the max load and controlling AISC load combination

max load factored load = ( Pa = 18 + 2 ) = 20 kips

attached below

4 0

3 years ago

Air enters a 200 mm diameter adiabatic nozzle at 195 deg C, 500 kPa and 100 m/s. It exits at 85 kPa. If the exit diameter is 158

ahrayia [7]

Answer:

$v_2 = 160.23 m/s$

$T_2 = 475.797 k$

Explanation:

given data:

Diameter = $d_1 = 200mm$

$t_1 =195 degree$

$p_1 =500 kPa$

$v_1 = 100m/s$

$p_2 = 85kPa$

$d_2 = 158mm$

from continuity equation

$A_1v_1 = A_2v_2$

$v_2 = \frac{\frac{\pi}{4}d_1^2 v_1^2}{\frac{\pi}{4}d_2^2}$

$v_2 = \frac{d_2v_1}{d_2^2}$

$v_2 = [\frac{d_1}{d_2}]^2 v_1$

$= [\frac{0.200}{0.158}]^2 \times 100$

$v_2 = 160.23 m/s$

by energy flow equation

$h_1 + \frac{v_1^2}{2} +gz_1 +q =h_2 + \frac{v_2^2}{2} +gz_2 +w$

$z_1 =z_2$ and q =0, w =0 for nozzle

therefore we have

$h_1 -h_2 =\frac{v_1^2}{2} -\frac{v_2^2}{2}$

$dh = \frac{1}{2} (v_1^2 -v_2^2)$

but we know dh = Cp dt

hence our equation become

$Cp(T_2 -T_1) = \frac{1}{2} (v_1^2 -v_2^2)$

$Cp (T_2 -T_1) = 7836.94$

$(T_2 -T_1) = \frac{7836.94}{1.005*10^3}$

$(T_2 -T_1) = 7.797$

$T_2 = 7.797 +468 = 475.797 k$

8 0

3 years ago

How to comment other people

Mekhanik [1.2K]

What you mean?? .!.!.!.!!.

4 0

3 years ago

Read 2 more answers

A circular column is fixed at the base and not supported at the top. If the column needs to be 15ft and hold 10kips, what is the

muminat

Answer:

The required size of column is length = 15 ft and diameter = 4.04 inches

Explanation:

Given;

Length of the column, L = 15 ft

Applied load, P = 10 kips = 10 × 10³ Psi

End condition as fixed at the base and free at the top

thus,

Effective length of the column, $\L_e$ = 2L = 30 ft = 360 inches

now, for aluminium

Elastic modulus, E = 1.0 × 10⁷ Psi

Now, from the Euler's critical load, we have

$P =\frac{\pi^2EI}{L_e^2}$

where, I is the moment of inertia

on substituting the respective values, we get

$10\times10^3 =\frac{\pi^2\times1.0\times10^7\times I}{360^2}$

I = 13.13 in⁴

also for circular cross-section

I = $\frac{\pi}{64}\times d^4$

thus,

13.13 = $\frac{\pi}{64}\times d^4$

d = 4.04 inches

The required size of column is length = 15 ft and diameter = 4.04 inches

3 0

3 years ago