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Artist 52 [7]
3 years ago
11

Increasing the concentration of greenhouse gases in Earth's atmosphere decreases the transparency of the atmosphere to infrared

light. decreases the transparency of the atmosphere to visible light. increases the transparency of the atmosphere to visible light. increases the transparency of the atmosphere to infrared light. Which of the following statements about global warming is NOT true? Computer models indicate that only human contributions to atmospheric carbon dioxide can account for recent global temperature rises Atmospheric carbon dioxide levels have risen dramatically over the past hundred years or so. Rises in global mean temperature tend to be closely associated with rises in atmospheric carbon dioxide concentrations. The global mean temperature in the modern era is the highest is has ever been
Physics
1 answer:
Artist 52 [7]3 years ago
5 0

Answer:

Decrease the transparency of the atmosphere

Explanation:

Different gases have characteristic absorption bands, so-called greenhouse gases have very high bands in the infrared region and small bands in the region of visible light.

   When the radiation  arrives to pass the greenhouse gases (CO²) and affects the earth's surface after several processes, the earth emits radiation in the infrared region, but this is absorbed by the gases, creating a net heating, since that the transparency of the atmosphere is  DECREASE for infrared radiation

Reviewing the correct answers is: Decrease the transparency of the atmosphere

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In a football game, a 90 kg receiver leaps straight up in the air to catch the 0.42 kg ball the quarterback threw to him at a vi
irinina [24]

To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where,

m_{1,2}= Mass of each object

v_{1,2} = Initial velocity of each object

v_f= Final Velocity

Since the receiver's body is static for the initial velocity we have that the equation would become

m_2v_2 = (m_1+m_2)v_f

(0.42)(21) = (90+0.42)v_f

v_f = 0.0975m/s

Therefore the velocity right after catching the ball is 0.0975m/s

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3 years ago
Which of the following is not an essential feature of scientific explanations?
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Answer:

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6. A pitcher throws a ball toward home plate 18.39 meters away. If the ball is traveling at a constant 40.0 m/s, how long does i
Kruka [31]
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3 years ago
Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f
elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

W_T+W_F=K_f-K_i (1)

Where:

W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to \frac{1}{2}mv^{2}, so if the initial velocity v_i is equal to zero, the initial kinetic energy K_i is equal to zero.

Then, replacing the values on the equation and solving for v_f, we get:

W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}

\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0
3 years ago
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.
astra-53 [7]

Answer:

a)

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

t1 = 0 This is the moment it jumped into the air

0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

5 0
3 years ago
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