I NEED HELP WITH THE LAST QUESTION I’L MARK U AS BRANLIEST

The trade winds are tropical winds that blow toward the Equator. Choose the statement that best explains why the trade winds blo

AveGali [126]

The best and most correct answer among the choices provided by your question is the fourth option or letter D. Trade winds blow towards the equator because t<span>he Equator receives the most heat energy.

</span>The surface air that flows from these subtropical high-pressure belts toward the Equator is deflected toward the west in both hemispheres by the Coriolis effect. These winds blow<span> predominantly from the northeast in the Northern Hemisphere and from the southeast in the Southern Hemisphere.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

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3 0

3 years ago

] A new coal-fired 750 MWe power plant with a thermal efficiency of 42% burns 9000 Btu/lb coal, which contains 1.1% sulfur. a. I

Valentin [98]

Answer:

The net emissions rate of sulfur is 1861 lb/hr

Explanation:

Given that:

The power or the power plant = 750 MWe

Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:

$\mathtt{=( 0.42\times 9000\times 1055.06) J}$

= 3988126.8 J

= 3.99 MJ

Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.

i.e.

The mass of the coal that is burned per sec $=\dfrac{750}{3.99}$

The mass of the coal that is burned per sec = 187.97 lb/s

The mass of sulfur burned $= \dfrac{1.1}{100} \times 187.97 \ lb/s$

= 2.067 lb/s

To hour; we have:

= 7444 lb/hr

However, If a scrubber with 75% removal efficiency is utilized,

Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr

= 0.25 × 7444 lb/hr

= 1861 lb/hr

Hence, the net emissions rate of sulfur is 1861 lb/hr

8 0

2 years ago

What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a

Paha777 [63]

$E = mgh + \frac{1}2} m v^{2} + \frac{1}{2} I \omega^{2} = mgh + \frac{1}2} m r ^{2} \omega ^{2} + \frac{1}{2} I \omega^{2}$

for a solid cylinder: $I = \frac{1}{2} m r^{2}$
for a hollow cylinder: $I = mr^{2}$

I will look at the case of a hollow cylinder:

$E = mgh + I \omega ^{2} = constant \\ \\ I = \frac{mgh}{ \omega^{2} }$

That is as far as i get.

7 0

3 years ago

Select the best answer for the question.

Aleonysh [2.5K]

Answer:

I think the answer will be water ,sorry if ik wrong

4 0

3 years ago

A toy car (0.50 kg) runs on a frictionless track and has an initial kinetic energy of 2.2 J, as the drawing shows. The numbers b

steposvetlana [31]

Answer:

Height will be equal to 0.4489 m

Explanation:

We have given mass of the toy m = 0.50 kg

Initial kinetic energy K = 2.2 J

We have to fond the height of the hill over which car roast

When car will roast the hill its kinetic energy will be converted into potential energy and at maximum height all kinetic energy will be converted into potential energy

So at maximum height $mgh=K$

Acceleration due to gravity $g=9.8m/sec^2$

So 0.50×9.8×h = 2.2

h = 0.4489 m

6 0

3 years ago