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Ivanshal [37]
4 years ago
11

After a chemical reaction, the atomic nuclei are _________.

Physics
1 answer:
sleet_krkn [62]4 years ago
4 0

After a chemical reaction, the atomic nuclei are unchanged. (C)


The nucleii don't know a thing about the outside world during

physical or chemical processes ... not until NUCLEAR things

happen.

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The index of refraction of light varies from color to color. True or False?
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The index of refraction of light varies from color to color. TRUE.
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3 years ago
Help me with this review question please.
QveST [7]

Answer:

K E=( mv²)/2

=(60×3.5²)/2

=367.5J

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3 years ago
What happens to light waves from a star as the star moves away from Earth?
AURORKA [14]
<h2>Answer: Light waves have a redshift due to the Doppler effect </h2>

The astronomer Edwin Powell Hubble observed several celestial bodies, and when obtaining the spectra of distant galaxies he observed the spectral lines were displaced towards the red (red shift), whereas the nearby galaxies showed a spectrum displaced to the blue.  

From there, Hubble deduced that the farther the galaxy is, the more redshifted it is in its spectrum. <u>The same happens with the stars and this phenomenom is known as the Doppler effect. </u>

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8 0
3 years ago
Physics {deceleration}
seropon [69]
Using kinematic equation, v^2 - u^2 = 2as. 5^2 - 3^2 = 2a x 16. a = 0.5m/s^2. So particle will deaccelerate at 0.5m/s^2. ( v = final velocity, u= initial velocity, a= acceleration, s= displacement.)
7 0
3 years ago
After coming down a slope, a 60-kg skier is coasting northward on a level, snowy surface at a constant 15 m&gt;s. Her 5.0-kg cat
Vinil7 [7]

To solve this exercise, it is necessary to apply the concepts of conservation of the moment especially in objects that experience an inelastic colposition.

They are expressed as,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where,

m_1= mass of the skier

m_2= mass of the cat

v_1 = initial velocity of skier

v_2 = initial velocity of cat

v_f= final velocity of both

Re-arrange to find V_f we have,

V_f = \frac{m_1v_1+m_2v_2}{(m_1+m_2)}

V_f = \frac{(60)(15)+(5)(-3.8)}{(60+5)}

V_f = 13.55m/s

Once the final velocity is found it is possible to calculate the change in kinetic energy, so

\Delta KE = KE_i-KE_f

\Delta KE = \frac{1}{2}(m_1v_1^2+m_2v^2_2)-\frac{1}{2}(m_1+m_2)v_f^2

\Delta KE = \frac{1}{2}((60)(15)^2+(5)(-3.8)^2)-\frac{1}{2}(60+5)(13.55)^2

\Delta KE = 819.1J

Therefore the amount of kinetic energy converted in to internal energy is 819J

3 0
3 years ago
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