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2 years ago

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What do fuel cells, batteries and, solar cells have in common A.the all produce static electricity B. they are all sources of di

rect current C. they are all sources of alternating current D. can all be used to power automobiles

2 answers:

Svetlanka [38]2 years ago

4 0

B and D are both true.
A and C are both false.

Vaselesa [24]2 years ago

3 0

The correct answer is B :)

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Betelgeuse is the bright red star representing the left shoulder of the constellation Orion. All the following statements about

Sav [38]

Answer:

Option A; ITS SURFACE IS COOLER THAN THE SURFACE OF THE SUN.

Explanation:

A red supergiant star is a larger and brighter type of red giant star. Red supergiants are often variable stars and are between 200 to 2,000 times bigger than the Sun. Example is Betelgeuse.

Betelgeuse is one of the largest known stars, it has a diameter of about 700 times the size of the Sun or 600 million miles, it emits almost 7,500 times as much energy as the Sun, it has a rather low surface temperature (6000F compared to the Sun's 10,000F); this means that it has a more cooler surface than the Sun's surface.

This low temperature also means that the star will appear orange-red in color, and the combination of size and temperature makes it a kind of star called a red super giant.

Although, all the statements above are correct, the only one that can be inferred from the red color of Betelgeuse is that ITS SURFACE IS COOLER THAN THE SURFACE OF THE SUN.

3 0

2 years ago

The Lyman series comprises a set of spectral lines. All of these lines involve a hydrogen atom whose electron undergoes a change

mihalych1998 [28]

Answer:

a) 1.2*10^-7 m

b) 1.0*10^-7 m

c) 9.7*10^-8 m

d) ultraviolet region

Explanation:

To find the different wavelengths you use the following formula:

$\frac{1}{\lambda}=R_H(1-\frac{1}{n^2})$

RH: Rydberg constant = 1.097 x 10^7 m^−1.

(a) n=2

$\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(2)^2})=8227500m^{-1}\\\\\lambda=1.2*10^{-7}m$

(b)

$\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(3)^2})=9751111,1m^{-1}\\\\\lambda=1.0*10^{-7}m$

(c)

$\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(4)^2})=10284375m^{-1}\\\\\lambda=9.7*10^{-8}m$

(d) The three lines belong to the ultraviolet region.

8 0

2 years ago

On a horizontal frictionless surface a mass M is attached to two light elastic strings both having length l and both made of the

EastWind [94]

Answer:

ω = √(2T / (mL))

Explanation:

(a) Draw a free body diagram of the mass. There are two tension forces, one pulling down and left, the other pulling down and right.

The x-components of the tension forces cancel each other out, so the net force is in the y direction:

∑F = -2T sin θ, where θ is the angle from the horizontal.

For small angles, sin θ ≈ tan θ.

∑F = -2T tan θ

∑F = -2T (Δy / L)

(b) For a spring, the restoring force is F = -kx, and the frequency is ω = √(k/m). (This is derived by solving a second order differential equation.)

In this case, k = 2T/L, so the frequency is:

ω = √((2T/L) / m)

ω = √(2T / (mL))

6 0

2 years ago

Calculate the critical angle between glass (n = 1.90) and ice (n = 1.31)? 43° 52° 60° 75°

Ket [755]

Answer: $43 \°$

Explanation:

According to <u>Snell’s Law</u>:

$n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$

Where:

$n_{1}=1.90$ is the first medium index of refraction (glass)

$n_{2}=1.31$ is the second medium index of refraction (ice)

$\theta_{1}$ is the angle of the incident ray

$\theta_{2}$ is the angle of the refracted ray

In this context, the index of refraction is a number that describes how fast light propagates through a medium or material.

Now, the critical angle $\theta_{c}$ is the angle from which there is no refraction and all the incident light is reflected to the same medium from which it proceeds, that is, the total internal reflection occurs. This is only possible when the index of refraction of the medium where the light strikes is higher than the index of refraction of the other medium, then the second angle (the exit angle) will reach the $90\º$ , for this critical incident angle $\theta_{c}$ .

Since $n_{1}>n_{1}$ , $\theta_{1}=[tex]\theta_{c}$ [/tex] and $\theta_{2}=90\º$ , hence:

$n_{1}sin(\theta_{c})=n_{2}sin(90\º)$

$(1.90)sin(\theta_{c})=(1.31)sin(90\º)$

Isolating $\theta_{c}$ :

$\theta_{c}=sin^{-1} (\frac{1.31}{1.90})$

Finally:

$\theta_{c}=43.5\º$

The option that is close to this value is $43\º$

4 0

2 years ago

Kate is researching air pollution and finds some information on ozone. She knows that ozone is a good thing as part of the ozone

jek_recluse [69]

"At ground level, ozone contributes to smog" so it is also an air pollutant.

Option: A

<u>Explanation</u>:

ozone is naturally present in stratosphere and acts as shield against harmful ultraviolet radiations. But it acts a pollutant contributing to global warming when it is present in lower level atmosphere particularly troposphere. In this level it combines with primary pollutants that is "nitrogen oxides" and "volatile organic" compounds to form secondary pollutant which absorbs outgoing radiation and contributes in raising the temperature. It has harmful impacts on vegetation as well as human health.

7 0

2 years ago