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sergij07 [2.7K]
2 years ago
7

The burning of fossil fuels contributes to the addition of greenhouse gases to the atmosphere. These gases trap thermal energy i

n the Earth’s atmosphere. Which of the following would be the most reasonable changes that could result from an increase in the burning of fossil fuels?
A. The air all over the globe would become dirtier.

B. There would be a global rise in temperatures.

C. Ecosystems would become more diverse.

D. Rates of photosynthesis would decrease.
Physics
2 answers:
balandron [24]1 year ago
7 0
Hey there!

The answer would be B. There would be a global rise in temperatures.

If fossil fuels contribute to the addition of greenhouse gases to the atmosphere and these same gases increase because of burning fossil fuels then there would be a global rise in temperatures. Which is answer B.

Hope this helps!
denpristay [2]2 years ago
5 0

Answer:

B. there would be a global rise in temperatures

Hence, global warming

Explanation:

hope this helped! :D

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dopasuj wartości pracy z ramki do przedstawionych sytuacji ,a nastepnie wyraż te prace w dżulach uwaga jedna wartośc pracy nie b
Eduardwww [97]

Answer:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Explanation:

Całą część pytania można znaleźć na poniższym schemacie.

Z diagramu załączonego poniżej; mamy

A - samolot lotniczy

B - dźwig

C - podnoszenie ciężarów

Wszyscy to wiemy ;

1kJ = 1000 J

1MJ = 1000000 J

Mamy cztery opcje; i.e 200 MJ, 800 kJ, 1.6 kJ  and 250 mJ

Z czterech opcji można wykluczyć 250 mJ, ponieważ jest to 0,25 J, co przedstawia bardzo niską energię w porównaniu z trzema warunkami pokazanymi na schemacie.

Więc:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Największą pracę wykona samolot. Jest tak, ponieważ ma bardzo dużą masę i bardzo dużą prędkość. W związku z tym istnieje potrzeba wytworzenia ogromnej ilości ciepła i energii.

Z drugiej strony żuraw może podnieść ładunek o wiele większy i przewyższa ciężar ciężaru, więc praca wykonywana przez dźwig musi być zdecydowanie większa niż praca ciężarka.

3 0
1 year ago
A 1.00 kg ball traveling towards a soccer player at a velocity of 5.00 m/s rebounds off the soccer
matrenka [14]

Answer:

A)   F = - 8.5 10² N,  B)   I = 21 N s

Explanation:

A) We can solve this problem using the relationship of momentum and momentum

          I = Δp

in this case they indicate that the body rebounds, therefore the exit speed is the same in modulus, but with the opposite direction

         v₀ = 8.50 m / s

         v_f = -8.50 m / s

         F t = m v_f -m v₀

         F = m \frac{(v_f - v_o)}{t}

let's calculate

         F = 1.00 \ \frac{(-8.5-8.5)}{2 \ 10^{-2}}

         F = - 8.5 10² N

B) let's start by calculating the speed with which the ball reaches the ground, let's use the kinematic relations

         v² = v₀² - 2g (y- y₀)

as the ball falls its initial velocity is zero (vo = 0) and the height upon reaching the ground is y = 0

         v = \sqrt{2g y_o}

calculate  

         v = \sqrt{2 \ 9.8 \ 10}

         v = 14 m / s

to calculate the momentum we use

         I = Δp

         I = m v_f - mv₀

when it hits the ground its speed drops to zero

we substitute

         I = 1.50 (0-14)

         I = -21 N s

the negative sign is for the momentum that the ground on the ball, the momentum of the ball on the ground is

        I = 21 N s

4 0
1 year ago
somewhere between the earth and the moon is a point where the gravitational attraction of the earth is canceled by the gravitati
mote1985 [20]
<span>It's pretty easy problem once you set it up.

Earth------------P--------------Moon

"P" is where the gravitational forces from both bodies are acting equally on a mass m

Let's define a few distances.
Rep = distance from center of earth to P
Rpm = distance from P to center of moon
Rem = distance from center of earth to center of moon

You are correct to use that equation. If the gravitational forces are equal then

GMearth*m/Rep² = Gm*Mmoon/Rpm²

Mearth/Mmoon = Rep² / Rpm²

Since Rep is what you're looking for we can't touch that. We can however rewrite Rpm to be

Rpm = Rem - Rep

Mearth / Mmoon = Rep² / (Rem - Rep)²

Since Mmoon = 1/81 * Mearth
81 = Rep² / (Rem - Rep)²

Everything is done now. The most complicated part now is the algebra, so bear with me as we solve for Rep. I may skip some obvious or too-long-to-type steps.

81*(Rem - Rep)² = Rep²
81*Rep² - 162*Rem*Rep + 81*Rem² = Rep²
80*Rep² - 162*Rem*Rep + 81*Rem² = 0

We use the quadratic formula to solve for Rep:
Rep = (81/80)*Rem ± (9/80)*Rem
Rep = (9/8)*Rem and (9/10)*Rem

Obviously, point P cannot be 9/8 of the way to the moon because it'll be beyond the moon. Therefore, the logical answer would be 9/10 the way to the moon or B.

Edit: The great thing about this idealized 2-body problem, James, is that it is disguised as a problem where you need to know a lot of values but in reality, a lot of them cancel out once you do the math. Funny thing is, I never saw this problem in physics during Freshman year. I saw it orbital mechanics in my junior year in Aerospace Engineering. </span> sylent_reality · 8 years ago
8 0
2 years ago
You slip a wrench over a bolt. Taking the origin at the bolt, the other end of the wrench is at x=18cm, y=5.5cm. You apply a for
mart [117]

Answer:

The torque on the wrench is 4.188 Nm

Explanation:

Let r = xi + yj where is the distance of the applied force to the origin.

Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,

r = 0.18i + 0.055j

The applied force f = 88i - 23j

The torque τ = r × F

So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j

= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j  

= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0   since i × i = 0, j × j = 0, i × j = k and j × i = -k

= 0 - 4.14k + 0.0484(-k) + 0

= -4.14k - 0.0484k

= -4.1884k Nm

≅ -4.188k Nm

So, the torque on the wrench is 4.188 Nm

8 0
2 years ago
A car drives 16 miles south and then 12 miles west. What is the magnitude of the car’s displacement? 4 miles 16 miles 20 miles 2
Ostrovityanka [42]
For this case, what we can do is use the Pythagorean theorem to find the magnitude of the displacement of the car.
 We have then
 d ^ 2 = 16 ^ 2 + 12 ^ 2&#10;
 From here, we clear the value of d.
 We have then:
 d =  \sqrt{16 ^ 2 + 12 ^ 2} &#10;
 Rewriting:
 d = \sqrt{256 + 144}
 d = \sqrt{400}
 d = 20 miles&#10;
 Answer:
 
The magnitude of the car's displacement is:
 
d = 20 miles
7 0
2 years ago
Read 2 more answers
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