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Oliga [24]
2 years ago
13

Carbon Dioxide levels fluctuate in part due to which of the following?

Physics
1 answer:
Ostrovityanka [42]2 years ago
8 0

Answer:

The life cycle of plants

Explanation:

Plant take in carbon dioxide for photosynthesis and release it when they die in the form of coal

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7. You start to walk toward the east towards home at a constant speed of 4 km/hr. At the same time, Someone else leaves your hom
amm1812

A) position time graph for both is shown

here one of the graph is of lesser slope which means it is moving with less speed while other have larger slope which shows larger speed

At one point they intersects which is the point where they both will meet

B) Let the two will meet after time "t"

now we can say that

if they both will meet after time "t"

then the total distance moved by you and other person will be same as the distance between you and home

so it is given as

v_1t + v_2t = d

4*t + 28*t = 3.2 km

t = \frac{3.2}{32} = 0.1 hr

so they will meet after t = 6 min

so from position time graph we can see that two will meet after t = 6 min where at this position two graphs will intersect


4 0
3 years ago
A Carnot engine operates with an efficiency of 26.0% when the temperature of its cold reservoir is 281 K. Assuming that the temp
VLD [36.1K]

Answer:

The temperature of cold reservoir should be 246.818 K for efficiency of 35%

Explanation:

In first case we have given efficiency of Carnot engine = 26 % = 0.26

Temperature of cold reservoir T_L=281K

We know that efficiency of Carnot engine is given by

\eta =1-\frac{T_L}{T_H}

0.26 =1-\frac{281}{T_H}

T_H=379.72K

For second Carnot engine efficiency is given as 35% = 0.35

And temperature of hot reservoir is same so T_H=379.72K

So 0.35=1-\frac{T_L}{379.72}

T_L=246.818K

So the temperature of cold reservoir should be 246.818 K for efficiency of 35%

4 0
3 years ago
Use the drop-down menu to complete the statement. Based on the field lines, the electric charges indicated by the question marks
zheka24 [161]

Answer: The same

Explanation: I just did it on Edg

5 0
3 years ago
Read 2 more answers
A mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 30. cm3. If the combustion of this mixtu
Y_Kistochka [10]

Answer:

Explanation:

Given

Original volume V1=30cm^3 converting to L

=30/1000=0.03L

Constant pressure P= 648 tors

Converting to atm; 648 tors*1atm/760 torr=0.853 atm

Work=984J= 984**1L/101.33=9.7L.atm

Note before

W= -P(Vfinal-Vinitial)

-9.7/0.853+0.03L=11.68L

5 0
3 years ago
Read 2 more answers
At highway speeds a car can accelerate at 1.7 m/s2.
Vaselesa [24]

Answer: 2.5 seconds

Explanation:

We know that the acceleration is:

a(t) = 1.7 m/s^2

To get the velocity function, we must integrate over time, and we will get:

v(t) = (1.7m/s^2)*t + v0

Where v0 is the initial velocity, in this case, we assume that we start at 23.6m/s, then the initial velocity is:

v0 = 23.6 m/s

Then the velocity equation is:

v(t) =  (1.7m/s^2)*t + 23.6 m/s

Now we want to find the value of t such v(t) = 27.8 m/s

Then:

v(t) =  27.8 m/s =  (1.7m/s^2)*t + 23.6 m/s

        27.8 m/s  - 23.6 m/s =  (1.7m/s^2)*t

         4.2 m/s = (1.7m/s^2)*t

         4.2m/s/(1.7m/s^2) = t = 2.5 s

Then at that acceleration, you need 2.5 seconds.

6 0
2 years ago
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