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vesna_86 [32]
3 years ago
5

Which describes the greenhouse effect? 1. Earth continues to get warmer over time because it is slowly moving closer to the sun.

2. Earth's atmosphere blocks the sun's ultraviolet light. 3. Pollution gets into the air, causing trees to not get enough oxygen. 4. Energy given off by earth is reflected off of earth's atmosphere back down to the surface.
Physics
1 answer:
Alekssandra [29.7K]3 years ago
6 0

Answer: The green house effect is best described by option 4 (Energy given off by earth is reflected off of earth's atmosphere back down to the surface).

Explanation:

The green house effect can be described as the energy given off by earth is reflected off of earth's atmosphere back down to the surface.

When energy from the sun passes through the atmosphere, some are absorbed which keeps the earth surface warm. While the rest is reflected back largely by cloud.

The energy which is emitted from the earth surface is called the infrared radiation. Some of the infrared radiation passess through the atmosphere but most is absorbed and re- emitted in all directions by the greenhouse gas molecules and clouds. This effect warms the earth surface and the lower atmosphere. Therefore this statement (Energy given off by earth is reflected off of earth's atmosphere back down to the surface) is correct about greenhouse effect.

For the greenhouse effect to occur, greenhouse gas molecules are mostly needed. Examples of these gases include:

--> Carbon dioxide (CO2),

--> Water vapor (H2O), and

--> Methane (CH4)

Over the years, the excessive human activities has lead to increase in the greenhouse gas molecules which has negatively affected the greenhouse effects.

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A planet has two
lozanna [386]
Kepler's third law hypothesizes that for all the small bodies in orbit around the
same central body, the ratio of (orbital period squared) / (orbital radius cubed)
is the same number.

<u>Moon #1:</u>  (1.262 days)² / (2.346 x 10^4 km)³

<u>Moon #2:</u>  (orbital period)² / (9.378 x 10^3 km)³

If Kepler knew what he was talking about ... and Newton showed that he did ...
then these two fractions are equal, and may be written as a proportion.

Cross multiply the proportion:

(orbital period)² x (2.346 x 10^4)³ = (1.262 days)² x (9.378 x 10^3)³

Divide each side by (2.346 x 10^4)³:

(Orbital period)² = (1.262 days)² x (9.378 x 10^3 km)³ / (2.346 x 10^4 km)³

               =  0.1017 day²

Orbital period = <u>0.319 Earth day</u> = about 7.6 hours.
7 0
3 years ago
A positive point charge (q = +8.65 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 2.97 m. A p
ella [17]

Answer:

r_B = 1.88 m

Explanation:

As we know that work done by electric force is given as

W_e = -q\Delta V

so here we know that charge is moving from

r_A = 2.97 m

to another position

so we will have

W_{AB} = \frac{kq_1q_2}{r_A} - \frac{kq_1q_2}{r_B}

-6.95\times 10^{-9} = (9\times 10^9)(8.65\times 10^{-8})(4.56\times 10^{-11})(\frac{1}{2.97} - \frac{1}{r_B})

-0.196 = (\frac{1}{2.97} - \frac{1}{r_B})

r_B = 1.88 m

7 0
3 years ago
The downsprue leading into the runner of a certain mold has a length of 175 mm. The cross-sectional area at the base of the spru
adoni [48]

Answer:

(a) Velocity at bottom is 1.85 m/s

(b) Volume flow rate is 7.4 x 10⁻⁴ m³/s.

(c) The time required to fill the mold is 1.35 s.

Explanation:

(a)

Applying Bernoulli's Equation on both ends of the down sprue, with the assumptions that every point is at atmospheric pressure and the liquid metal at the pouring basin is at zero velocity. The equation then becomes:

V = √2gh

where,

V = velocity at bottom of down sprue

h = height of down sprue = 175 mm = 0.175 m

V = √2(9.8 m/s²)(0.175 m)

<u>V = 1.85 m/s</u>

<u></u>

(b)

The volume flow rate is given as:

Volume Flow Rate = (V)(A)

where,

V = velocity at bottom = 1.85 m/s

A = Area of bottom = 400 mm² = 0.0004 m²

Therefore,

Volume Flow Rate = (1.85 m/s)(0.0004 m²)

<u>Volume Flow Rate = 7.4 x 10⁻⁴ m³/s = 740 cm³/s</u>

(c)

The time required to fill the cavity is given as:

Volume Flow Rate = V/t

where,

V = Volume of mold Cavity = 0.001 m³

t = time required to fill the cavity = ?

Therefore,

t = V/Volume Flow Rate

t = 0.001 m³/7.4 x 10⁻⁴ m³/s

<u>t = 1.35 s</u>

<u></u>

5 0
3 years ago
Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge
kramer

Answer:

E = 440816.32 N/C

Explanation:

Given data:

Three point charge of charge equal to +3.0 micro coulomb

fourth point charge = - 3.0 micro coulomb

side of square = 0.50 m

K =1/4 \pi \epsilon_0 = 8.99 \times 10^9 N.m^2/c^2

Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value

So we have

E_1 + E_3 = 0

E =E_2 + E_4

E = 2 E_2

[E_2 =\frac{2\times k \times q}{r^2}

[r= \frac{(0.5^2 + 0.5^2)^2}{2} = 0.35 m]

plugging all value

E = 2 E_2

E = 2 E_2 =\frac{2\times k \times q}{r^2}

E = \frac{2 \times 8.99 \times 10^93\times 10^{-6}}{0.35^2}

E = 440816.32 N/C

3 0
3 years ago
Read 2 more answers
Arrange the distances between Earth and various celestial objects in order from least to greatest. Use the conversion table to h
Kaylis [27]

distance to the star Betelgeuse: 640 ly

As we know that

1 ly = 63000 AU

also we know that

1AU = 1.5 \times 10^8 km

1 ly = 63000 (1.5 \times 10^8) = 9.45 \times 10^{12} km

So the distance of Betelgeuse = 640 ly

d_1 = 640 \times 9.45 \times 10^{12} = 6.05 \times 10^{15} m

distance to the star VY Canis Majoris: 3.09 × 10^8 AU

d_2 = 3.09\times 10^8 \times 1.5 \times 10^8 km

d_2 = 4.64 \times 10^{16} km

distance to the galaxy Large Magellanic Cloud: 49976 pc

1 pc = 3.262 ly = 3.262 \times 9.45 \times 10^{12} km

1pc = 3.08 \times 10^{13} km

now we have

d_3 = 49976 \times 3.08 \times 10^{13}

d_3 = 1.54 \times 10^{18} km

distance to Neptune at the farthest: 4.7 billion km

d_4 = 4.7 \times 10^9 km

now the order of distance from least to greatest is as following

1. distance to Neptune at the farthest

2. distance of Betelgeuse

3. distance to the star VY Canis Majoris

4. distance to the galaxy Large Magellanic Cloud

6 0
3 years ago
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