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2 years ago

Which of the following, if eliminated, would completely prevent the greenhouse effect from occurring on earth?

1 answer:

5 0

The substance, if eliminated, that would completely prevent the greenhouse effect from occurring on earth is fossil fuels (option C).

<h3>What is greenhouse effect?</h3>

Greenhouse effect is the process by which a planet is warmed by its atmosphere.

Greenhouse effect is caused by greenhouse gases that deplete the ozone layer.

Fossil fuels, when burned, will release greenhouse gases such as carbon etc. to the atmosphere.

Therefore, the substance, if eliminated, that would completely prevent the greenhouse effect from occurring on earth is fossil fuels.

Learn more about greenhouse gases at: brainly.com/question/14131369

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is it possible to breed a black haired long tailed retriever? If so, what percentage is possible? PLEASE answer the corrects ans

Pachacha [2.7K]

Answer: it can be considered a genetic mutation with a history of a Golden Retriever in their blood but it is very rare. and there our some black retrievers you can buy too. i hope i helped

Explanation:

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2 years ago

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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:
$C_s = \frac{1495 J}{(351 g)(66.0^{\circ}C-55.0^{\circ}C)}=0.39 J/g^{\circ}C$

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3 years ago

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If an object is placed between a convex lens and its focal point, which type of image will be produced?

Andreas93 [3]

<span>virtual, upright, and magnified</span>

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3 years ago

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Helen [10]

Answer:

wavelength decreases and frequency increase

Explanation:

the higher the wavelength the smaller the frequency , the smaller the wavelength the higher the frequency

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3 years ago

Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m

alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = $0.407\times 10^{-3}\ m$

Energy stored in the capacitor (U) = $7.87\ nJ=7.87\times 10^{-9}\ J$

Assuming the medium to be air.

So, permittivity of space (ε) = $8.854\times 10^{-12}\ F/m$

Area of the square plates is given as:

$A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2$

Capacitance of the capacitor is given as:

$C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F$

Now, we know that, the energy stored in a parallel plate capacitor is given as:

$U=\frac{CE^2d^2}{2}$

Rewriting in terms of 'E', we get:

$E=\sqrt{\frac{2U}{Cd^2}}$

Now, plug in the given values and solve for 'E'. This gives,

$E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C$

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0

3 years ago