Answer:
v= 449.8 m/s
Explanation:
Given data
Frequency= 346Hz
Wave length= 1.4m
The expression below is used to find the speed
substitute
Hence the speed is v= 449.8 m/s
Question 6 of 10
1 answer:
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Answer:
ee that the lens with the shortest focal length has a smaller object
Explanation:
For this exercise we use the constructor equation or Gaussian equation
where f is the focal length, p and q are the distance to the object and the image respectively.
Magnification a lens system is
m = = -
h ’= -\frac{h q}{p}
In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞
Let's calculate the distance to the image for each lens
f = 6.0 cm
as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf
q = f = 6.0 cm
for the lens of f = 12.0 cm q = 12.0 cn
to find the size of the image we use
h ’= h q / p
where p has a high value and is the same for all systems
h ’= h / p q
Thus
f = 6 cm h ’= fo 6 cm
f = 12 cm h ’= fo 12 cm
therefore we see that the lens with the shortest focal length has a smaller object
Answer:
critical angle = 64.79°
ray will not penetrate into the water
Explanation:
given details
refractive index of oil n1 = 1.47
refractive index of water n2= 1.33
incident angle = 71.4 °
= sin^{-1}(\frac{1.33}{1.47}
critical angle = 64.79°
incident angle > critical angle = reflection
incident angle < critical angle = refraction
In this case, 71.4 > 64.79 therefore it will undergo total internal reflection
therefore ray will not penetrate into the water