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algol13
3 years ago
12

Which source is most likely an authority on the subject? Dr. Vale, a pediatrician, discusses electrical circuitry on her home-im

provement website. Jonah, an amateur cook, teaches his son how to make the best grilled cheese sandwiches. Susan, an architect, presents her hand-made quilts at the Quilting Society. Dr. Patel, a cardiologist, presents her findings on heart disease.
Physics
2 answers:
Pavlova-9 [17]3 years ago
7 0

Answer:

D.

Explanation:

I took the quiz

Naddika [18.5K]3 years ago
6 0
The correct answer is Dr. Patel, a cardiologist, presents her findings on heart disease.

A cardiologist is a heart doctor, so he would be the highest authority to speak about heart disease. The others are all amateurs in their businesses.
You might be interested in
gayle cooks a roast in her microwave oven. the klystron tube in the oven emits photons whose energy is 1.20 x 10^-3 ev. what are
AveGali [126]

Answer:

\lambda=1.03\times 10^{-3}\ m

Explanation:

Given that,

The energy of the microwave oven is 1.2\times 10^{-3}\ eV.

We need to find the wavelength of these photons.

1.2\times 10^{-3}\ eV=1.2\times 10^{-3}\times 1.6\times 10^{-19}\\\\=1.92\times 10^{-22}\ J

The energy of a wave is given by :

E=\dfrac{hc}{\lambda}\\\\\lambda=\dfrac{hc}{E}

Put all the values,

\lambda=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{1.92\times 10^{-22}}\\\\\lambda=1.03\times 10^{-3}\ m

So, the wavelength of these photon is 1.03\times 10^{-3}\ m.

7 0
3 years ago
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
If you wanted to find the area of the hot filament in a light bulb, you would have to know the temperature (determinable from th
Sladkaya [172]

Answer:

To find out the area of the hot filament of a light bulb, you would need to know the temperature, the power input, the Stefan-Boltzmann constant and <u>Emissivity of the Filament</u>.

Explanation:

The emissive power of a light bulb can be given by the following formula:

E = σεAT⁴

where,

E = Power Input or Emissive Power

σ = Stefan-Boltzmann constant

ε = Emissivity

A = Area

T = Absolute Temperature

Therefore,

A = E/σεT⁴

So, to find out the area of the hot filament of a light bulb, you would need to know the temperature, the power input, the Stefan-Boltzmann constant and <u>Emissivity of the Filament</u>.

3 0
3 years ago
Rachel collected data on the speed of sound waves passing through different media.
Helga [31]

The conclusion is, medium Q is most likely a solid because solids have the highest density and sound waves travel fastest in high density media.

<h3>Effect of density on speed of sound</h3>

Sound wave is mechanical wave that requires material medium for its propagation.

A high dense medium, is a medium with closely packed molecules. Since sound wave requires material medium for its propagation, it will travel faster in a high dense medium than a less dense medium.

Thus, the speed of sound increases as the density of the medium increases.

<h3>Speed of sound in the different media</h3>

The conclusion that can be made from the speed of sound in the different media is "Medium Q is most likely a solid because solids have the highest density and sound waves travel fastest in high density media".

Learn more about effect of density on speed of sound here: brainly.com/question/3323620

7 0
2 years ago
Which statement is true about an object moving in a circular motion due to centripetal force, F, when the mass is doubled?
marshall27 [118]

Ans) A) Centripetal force will be doubled.

See centripetal force F = mv^2/r

That means centripetal force is directly proportional to the mass of the particle

So, if we double the mass, centripetal force will be increased by twofolds.

So, option A) is correct.

Now, looking at the other options,

B) says centripetal force is unaltered which is incorrect as centripetal force has been altered and increased twofold.

Option C) and D) reduces centripetal force which are also not possible here.

So, only Option A) is correct

4 0
3 years ago
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