Answer:
If the frequency of the motion of a simple harmonic oscillator is doubled , then maximum speed of the oscillator changes by the factor 2
Explanation:
We know that in a simple harmonic oscillator the maximum speed is given by
= 
Here A is amplitude which is constant , so from above equation we see that maximum speed is directly proportional to 
of the oscillation .
Since 
= 2
Where 
is the maximum speed when frequency is doubled .
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Answer:
The speed of the ball was, v = 3 m/s
Explanation:
Given data,
The time period of the ball, t = 8 s
The distance the ball rolled, d = 24 m
The velocity of an object is defined as the object's displacement to the time taken. The formula for the velocity is,
v = d / t m/s
Substituting the given values in the above equation,
v = 24 / 8
= 3 m/s
Hence, the speed of the ball was, v = 3 m/s
Answer:
The correct answer to that question is HENRY
Answer:
A) d = 11.8m
B) d = 4.293 m
Explanation:
A) We are told that the angle of incidence;θ_i = 70°.
Now, if refraction doesn't occur, the angle of the light continues to be 70° in the water relative to the normal. Thus;
tan 70° = d/4.3m
Where d is the distance from point B at which the laser beam would strike the lakebottom.
So,d = 4.3*tan70
d = 11.8m
B) Since the light is moving from air (n1=1.00) to water (n2=1.33), we can use Snell's law to find the angle of refraction(θ_r)
So,
n1*sinθ_i = n2*sinθ_r
Thus; sinθ_r = (n1*sinθ_i)/n2
sinθ_r = (1 * sin70)/1.33
sinθ_r = 0.7065
θ_r = sin^(-1)0.7065
θ_r = 44.95°
Thus; xonsidering refraction, distance from point B at which the laser beam strikes the lake-bottom is calculated from;
d = 4.3 tan44.95
d = 4.293 m
