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3 years ago

Earth's atmosphere contains only small amounts of carbon dioxide because

1 answer:

4 0

Of biogeochemical cycles

Because certain biogeochemical cycles take place to lessen these gases from forming since they are not beneficial for life if too much.
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Sir wants to see if a new shower cleaner works better in removing soap dirt than his old cleaner. He uses the new cleaner on on

sergeinik [125]

The independent variable in this problem would be the different types of shower cleaner. The dependent variable would be the shower tiles.

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3 years ago

A puddle of water has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavel

erastovalidia [21]

Answer:

Explanation:

To solve the problem it is necessary to apply the concepts of Destructive and constructive interference. The constructive interference in tin film is given by

$2t = (m+\frac{1}{2})\frac{\lambda}{n}$

Where,

t = thickness

$\lambda=$ Wavelenght

m= is an integer

n= film/refractive index

We use this equaton because phase change is only present for gasoline air interface, but not at the gasoline-water interface. <em>The minimum t only would be when the value of m=0 then</em>

$2nt = \frac{\lambda}{2}$

$t = \frac{560nm}{4*1-4}$

$t = 100nm$

Therefore the correct answer is D. The minimum thickness of the film to see ab right reflection is 100nm

4 0

3 years ago

A young kid of mass m = 36 kg is swinging on a swing. The length from the top of the swing set to the seat is L = 3.5 m. The boy

lara31 [8.8K]

Answer:

Explanation:

Given

mass of boy=36 kg

length of swing=3.5 m

Let T be the tension in the swing

At top point $mg-T=\frac{mv^2}{r}$

where v=velocity needed to complete circular path

Th-resold velocity is given by $mg-0=\frac{mv^2}{r}$

$v=\sqrt{gr}=\sqrt{9.8\times 3.5}=5.85 m/s$

So apparent weight of boy will be zero at top when it travels with a velocity of $v=\sqrt{gr}$

To get the velocity at bottom conserve energy at Top and bottom

At top $E_T=mg\times 2L+\frac{mv^2}{2}$

Energy at Bottom $E_b=\frac{mv_0^2}{2}$

Comparing two as energy is conserved

$v_0^2=4gl+gl$

$v_0^2=5gL$

$v_0=\sqrt{5gL}=13.09 m/s$

Apparent weight at bottom is given by

$W=\frac{mv_0^2}{L}-mg=\frac{36\times 13.09^2}{3.5}+36\times 9.8=2115.23 N$

6 0

3 years ago

Relate a real life phenomenon with each branch of physics

anastassius [24]

Answer:

Branches of physics with real life examples

In measuring and understanding nuclear fission (a real life phenomenon), all branches of theoretical and experimental physics have to be employed. Physics branches needed in it are, radiation detection and measurement, nuclear physics, statistical physics, thermodynamics, and almost all others.

Explanation:

4 0

3 years ago

If 1495 j of heat is needed to raise the temperature of a 351 g sample of a metal from 55.0°c to 66.0°c, what is the specific he

forsale [732]

The amount of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q= mC_s \Delta T$
where m is the mass of the substance, Cs is its specific heat capacity and $\Delta T$ is the increase of temperature.

If we re-arrange the formula, we get
$C_s = \frac{Q}{m \Delta T}$
And if we plug the data of the problem into the equation, we can find the specific heat capacity of the substance:
$C_s = \frac{1495 J}{(351 g)(66.0^{\circ}C-55.0^{\circ}C)}=0.39 J/g^{\circ}C$

6 0

3 years ago

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