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3 years ago

Why would it be desirable to use an energy resource that does not produce greenhouse gases ?

2 answers:

8 0

It be desirable to use an energy resource that does not produce greenhouse gases because it contribute to climate change .

Option C

<u>Explanation:</u>

Greenhouse gases are the toxic gases that contribute to the "greenhouse effect". Greenhouse gases have the ability to absorb and emit some of the outgoing energy that is radiated by the Earth's surface, eventually causing the heat to get accumulated in the "lower atmosphere". It is always beneficial to use Renewable Energy sources such as Wind, solar and water which gives no harm to the environment and climate with no greenhouse gas emissions.

<u>For example:</u> If we have to generate electricity, there will be no greenhouse gas production at the beginning and a very less amount of production at the entire life cycle.

VikaD [51]3 years ago

7 0

Answer:

Explanation:

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A bat hits a moving baseball. If the bat delivers a net eastward impulse of 1.3 N-s and the ball starts with an initial horizont

Georgia [21]

Answer:

$141.30grams$

Explanation:

First, denote our known values;

$J=1.3N-s(+,east)\\u=-3.8ms^-^1(-,west)\\v=5.4ms^-^1(+,east)\\m=?$

Mass is impulse divided by change in velocity:

$m=\frac{J}{v-u}\\=\frac{1.3}{5.4--3.8}\\\\=\frac{1.3}{9.2}\\=0.1413Kgs$

Hence, the mass of the ball is 141.30grams

3 0

3 years ago

What are the effects of radiation on different surfaces such as an ice- covered lake, a forest, an ocean, or an asphalt road?

jarptica [38.1K]

Answer:

it could kill the fish in water

4 0

3 years ago

Read 2 more answers

. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of

Olegator [25]

Answer:

See Explanation

Explanation:

Given

$s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}$

Solving (a): Units and dimension of $s_0$

From the question, we understand that:

$s \to L$ --- length

$t \to T$ --- time

Remove the other terms of the equation, we have:

$s=s_0$

Rewrite as:

$s_0=s$

This implies that $s_0$ has the same unit and dimension as $s$

Hence:

$s_0 \to L$ --- dimension

$s_o \to$ Length (meters, kilometers, etc.)

Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

Make $v_0$ the subject

$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

$v_0 = LT^{-1}$

Hence:

$v_0 \to LT^{-1}$ --- dimension

$v_0 \to$ $m/s$ --- unit

Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

$a_0 \to$ $m/s^2$ --- acceleration

Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

Hence:

$S_0 = LT^{-4$ --- dimension

$S_0 \to$ $m/s^4$ --- unit

Solving (e): Units and dimension of $c$

Ignore other terms of the equation, we have:

$s=\frac{ct^5}{120}$

Rewrite as:

$\frac{ct^5}{120} = s$

Make $c$ the subject

$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

$c = \frac{L}{T^5}$

$c = LT^{-5}$

Hence:

$c \to LT^{-5}$ --- dimension

$c \to m/s^5$ --- units

4 0

3 years ago

Which telescopes must be placed in orbit around earth in order to observe short-wavelength radiation?.

son4ous [18]

Space telescopes must be placed in orbit around earth in order to observe short-wavelength radiation.

<h3>What is telescope?</h3>

A telescope is an optical instrument that uses lenses, curved mirrors, or a combination of both to watch distant objects.

When atoms in a gas reach this temperature, they travel so quickly that when they collide, they release X-ray photons with wavelengths smaller than 10 nanometers.

Because the Earth's atmosphere prevents all X-rays from space, these wavelengths must be seen using space telescopes.

To study short-wavelength radiation, space telescopes must be put in orbit around the Earth.

Hence, space telescope is the correct answer.

To learn more about the telescope, refer:

brainly.com/question/556195

#SPJ1

6 0

2 years ago

Which describes why people on earth can see light from the stars in the sky that are so far away?

Ugo [173]

Yes, C is correct. It self explains itself as we know light travels through a vacuum ( doesn't need a medium) and light is a type of electromagnetic wave.

7 0

3 years ago