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insens350 [35]
3 years ago
5

Select the correct answer. In a given chemical reaction, the energy of the products is greater than the energy of the reactants.

Which statement is true for this reaction? A. Energy is absorbed in the reaction. B. Energy is released in the reaction. C. No energy is transferred in the reaction. D. Energy is created in the reaction. E. Energy is lost in the reaction.
Physics
1 answer:
ioda3 years ago
3 0
B. Energy is released in the reaction... just looked it up!
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Write down the (real) electric and magnetic fields for a monochromatic plane wave of amplitude E0 , frequency w, and phase angle
jeka57 [31]

Answer:

a) the oscillation of this field is in phase, when the magnetic field goes in the negative direction of y, the elective field goes in the positive direction of the z axis

b) the direction of the magnetic field perpendicular to this electric field and the speed in the negative x the magnetic field goes in the x direction and in the direction (1, - 1.1)

Explanation:

a) the polarization the determined wave oscillates the electric field, which is the z axis

 As the wave travels on the negative x-axis and the magnetic field is perpendicular, this field goes on the positive y-axis

the oscillation of this field is in phase, when the magnetic field goes in the negative direction of y, the elective field goes in the positive direction of the z axis

be) in the case of a polarization in the xi plane the magnetic field must go in the direction of the magnetic field perpendicular to this electric field and the speed in the negative x the magnetic field goes in the x direction and in the direction (1, - 1.1)

3 0
3 years ago
The doppler effect is used to determine?
Korolek [52]
Uhh it is used to detirmine heat

7 0
3 years ago
Read 2 more answers
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
Explain in terms of energy flow how a cold pack works on a sprained ankle
sergij07 [2.7K]
The <span>flow of how a cold pack works on a sprained ankle is based on the second law of thermodynamics which states that energy will flow from a higher to a lower temperature. So your body heat will flow to the cold pack in which you will feel the coldness of the pack.</span>
7 0
3 years ago
A airplane travels 3260 kilometers in 4 hours. What is the airplane average speed?
zavuch27 [327]
I'm pretty sure that it's 815.
4 0
3 years ago
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