Ask question

Ask question

All categories

2 years ago

9

Although the temperature gradient changes from region to region in the homosphere, there is one gradient that stays the same. it

continues to decrease as you increase in altitude, no matter where you are in the homospere. what gradient?

1 answer:

bezimeni [28]2 years ago

3 0

Answer: the answer is air gradient

Explanation:

You might be interested in

A batter hits a baseball with a bat. The bat exerts a force on the ball. Does the ball exert a force on the bat?

Mazyrski [523]

Yes, think about the difference of swinging a bat and not hitting a ball. It's fairly easy right? Now, when you hit a ball with the bat, you will feel the bat sting your hands. That's the force the ball is exerting on the bat!

3 0

3 years ago

Read 2 more answers

What is the correct answer?

pentagon [3]

Answer:

B) x^2+6x+8

Explanation:

x-4 | x^3+2x^2-16x-32

- x^3-4x^2 <-- (x-4)(x^2)

_________________

6x^2-16x-32

- 6x^2-24x <-- (x-4)(6x)

_________________

8x-32

- 8x-32 <- (x-4)(8)

___________________________

0 | x^2+6x+8

This means the answer is B) x^2+6x+8

3 0

3 years ago

EXPLAIN HOW ENERGY IS TRANSMITTED THROUGH A MEDIUM

jek_recluse [69]

I say it helped then because TrueType had room

5 0

3 years ago

Read 2 more answers

What is newtons 3rd law? pls help!

drek231 [11]

For every actions, there is an opposite reaction.

3 0

3 years ago

Read 2 more answers

At a certain location, wind is blowing steadily at 9 m/s. Determine the mechanical energy of air per unit mass and the power gen

Misha Larkins [42]

Answer:

The specific mechanical energy of the air in the specific location is 40.5 J/kg.
The power generation potential of the wind turbine at such place is of 2290 kW
The actual electric power generation is 687 kW

Explanation:

The mechanical energy of the air per unit mass is the specific kinetic energy of the air that is calculated using: $\frac{1}{2} V^2$ where V is the velocity of the air.
The specific kinetic energy would be: $\frac{1}{2}(9\frac{m}{s})^2=40.5\frac{m^2}{s^2}=40.5\frac{m^2 }{s^2}\frac{kg}{kg}=40.5\frac{N*m }{kg}=40.5\frac{J}{kg}$ .
The power generation of the wind turbine would be obtained from the product of the mechanical energy of the air times the mass flow that moves the turbine.
To calculate mass flow it is required first to calculate the volumetric flow. To calculate the volumetric flow the next expression would be: $\frac{V\pi D_{blade}^2}{4} =\frac{9\frac{m}{s}\pi(80m)^2}{4} =45238.9\frac{m^3}{s}$
Then the mass flow is obtain from the volumetric flow times the density of the air: $m_{flow}=1.25\frac{kg}{m^3}45238.9\frac{m^3}{s}=56548.7\frac{kg}{s}$
Then, the Power generation potential is: $40.5\frac{J}{kg} 56548.7\frac{kg}{s} =2290221W=2290.2kW$
The actual electric power generation is calculated using the definition of efficiency: $\eta=\frac{E_P}{E_I}}$ , where η is the efficiency, $E_P$ is the energy actually produced and, $E_I$ is the energy input. Then solving for the energy produced: $E_P=\eta*E_I=0.30*2290kW=687kW$

6 0

3 years ago