Answer:
Option B, Some of the cars' kinetic energy was converted to sound and heat energy.
Explanation:
In an elastic collision, no energy is lost during and after collision. Thus, it can be said that in an elastic collision both momentum and kinetic energy remains conserved.
While in non-elastic collision, kinetic energy of the system is lost. However, the momentum of the system is conserved. Generally, during and after collision some of the kinetic energy is lost as thermal energy, sound energy etc.
Hence, option B is correct
Answer:
False, Sunspots appear dark (in visible light) due to their low temperature(cooler) than rest of the sun
Explanation:
Sunspots appear dark because they are much cooler( have low temperature than the rest of the surface contained by Sun. As they appear dark, but still they have very temperature that's why so hot. Sunspots have temperatures ranges 3,500 Celsius (3773 kelvin) and the surrounding surface of the sun has a temperature much higher of about 5,500 Celsius(5773 Kelvin). Even if we see a sunspot alone in space, it will glow so brightly.
Learn more about sunspots :
brainly.com/question/27774496
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Answer:
Photons.
Explanation:
Einstein explained the photoelectric effect by proposing that light is transmitted, not in the form of continuous waves, but in the form of many tiny, discrete packets of energy called photons. Photons are massless and chargeless particle. The energy of a photon is given by :

f is the frequency of photon.
C is correct answer.
Walking along a 6 meter beam without falling helps to develop balance.
Hope it helped.
-Charlie
Answer:
1.19 m/s²
Explanation:
The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so
f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀ (1)
Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²
Equating (1) and (2) we ave
2(√mg/μ)/f = T²g/4π²
Making g subject of the formula
g = 2π√(2√(m/μ)/f)/T
The period T = 316 s/100 = 3.16 s
Substituting the other values into , we have
g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16
g = 2π√(2 × 35.877/200 Hz)/3.16
g = 2π√(71.753/200 Hz)/3.16
g = 2π√(0.358)/3.16
g = 2π × 0.599/3.16
g = 1.19 m/s²