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2 years ago

Which of the following is true of greenhouse gases?

Physics

1 answer:

Sedaia [141]2 years ago

4 0

Answer:

A. They trap energy in the atmosphere.

Explanation:

Greenhouse gases are defined as gases that absorb and emit radiation within the infrared range. These gases allow the sun’s rays to pass through the ozone layer and warm the earth, but prevent this warmth from escaping atmosphere.

Greenhouse gases cause the greenhouse effect on planets. Examples of greenhouse gases in Earth's atmosphere are water vapor, carbon dioxide, methane, ozone etc.

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What are two examples of goods and two examples of services

julsineya [31]

The goods and the services make up the basis of every economy. The goods can simply be defined as merchandise or possessions. The services can be defined as the actions through which help is provided, or work is done for someone else. Example of goods are the food and furniture, with the food being crucial for the survival of the people, while the furniture is an essential part of every home and its practicality and decor. Examples of services are teaching and car repairing. The teaching is crucial for the development of the societies, as through it the people get education, while the repairing of cars is very important as lot of people have them, can not afford to buy new ones all the time, and they need for their daily movement over longer distances.

8 0

3 years ago

if the efficiency of an electric furnace is 96%, then 96% of the input energy is transformed into thermal energy. what is the ot

Nonamiya [84]

It is wasted, most likely as light, in this case, or it is lost during the transport of electricity.

5 0

3 years ago

Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid

Korolek [52]

Answer:

$z= 2.5 \ m$

$z = (1 \ m , 4 \ m )$

Explanation:

From the question we are told that

Their distance apart is $d = 5.00 \ m$

The wavelength of each source wave $\lambda = 6.0 \ m$

Let the distance from source A where the construct interference occurred be z

Generally the path difference for constructive interference is

$z - (d-z) = m \lambda$

Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0

$z - (5-z) = 0$

=> $2 z - 5 = 0$

=> $z= 2.5 \ m$

Generally the path difference for destructive interference is

$|z-(d-z)| = (2m + 1)\frac{\lambda}{2}$

=> $|2z - d |= (0 + 1)\frac{\lambda}{2}$

=> $|2z - d| =\frac{\lambda}{2}$

substituting values

$|2z - 5| =\frac{6}{2}$

=> $z = \frac{5 \pm 3}{2}$

$z = \frac{5 + 3}{2}$

$z = 4\ m$

and

$z = \frac{ 5 -3 }{2}$

=> $z = 1 \ m$

=> $z = (1 \ m , 4 \ m )$

7 0

3 years ago

A crane does 300 Joules of work in 6 seconds. How much power was used?

HACTEHA [7]

Answer: P= mad/t or P=w/t so P= 300/6= 50 W

6 0

3 years ago

A proton initially moves left to right along the x-axis at a speed of 2.00 x 103 m/s. It moves into an uniform electric field, w

FinnZ [79.3K]

Answer:

E = 1.04*10⁻¹ N/C

Explanation:

Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:

$vf^{2} -vo^{2} = 2*a*x$

As the proton is coming at rest after travelling 0.200 m to the right, vf = 0, and x = 0.200 m.

Replacing this values in the equation above, we can solve for a, as follows:

$a = \frac{vo^{2}*mp}{2*x} = \frac{(2.00e3m/s)^{2}}{2*0.2m} = 1e7 m/s2$

According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:

F = mp*a = q*E

For a proton, we have the following values:

mp = 1.67*10⁻²⁷ kg

q = e = 1.6*10⁻¹⁹ C

So, we can solve for E (in magnitude) , as follows:

$E = \frac{mp*a}{e} =\frac{1.67e-27kg*1e7m/s2}{1.6e-19C} = 1.04e-1 N/C$

⇒ E = 1.04*10⁻¹ N/C

5 0

3 years ago