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Len [333]
3 years ago
15

A 1.80-m-tall person stands 3.50 m from a convex mirror and notices that he looks precisely half as tall as he does in a plane m

irror placed at the same distance.
What is the radius of curvature of the convex mirror? (Assume that sinθ≈θ.) [Hint: The viewing angle is half.]
Physics
1 answer:
svlad2 [7]3 years ago
5 0

Answer:

Explanation:

Given:

height of the object h_o=1.90m

height of the image h_i=\frac{h_o}{2}\\=h_i=\frac{1.80}{2}=0.90m

object distance d_o=3.50m

we know that: \frac{h_i}{h_o}=\frac{-d_i}{d_o}\\\\=d_i=\frac{-h_i}{h_o}\times d_o\\\\=-\frac{0.90}{1.80}\times (3.50)=-1.75m

image d_i=-1.75m

According to lens formular:

\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}\\\\=\frac{1}{3.50}+-\frac{1}{1.75}\\\\f=-3.5

focal length=\frac{radius}{2}\\\\radius=2\times f\\=2\times 3.5=7m

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Answer:

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7 0
3 years ago
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Pani-rosa [81]
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3 0
4 years ago
What is the instantaneous speed at 5 seconds? (remember s=d/t)
TEA [102]
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Up until that time, the speed has been 1 m/s. And then, at exactly 5 seconds, it becomes zero.
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4 0
4 years ago
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A man 2 m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 23 m tall. When the man is 10 m from th
Evgen [1.6K]

Answer:

\dfrac{d\theta}{dt}=0.038\ rad/s

Explanation:

Given that

\dfrac{dx}{dt}= -1\ m/s

From the diagram

tan\theta=\dfrac{21}{x}

By differentiating with time t

sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}

When x= 10 m

tan\theta=\dfrac{21}{10}

θ = 64.53°

Now by putting the value in equation

sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}

sec^264.53^{\circ} \dfrac{d\theta}{dt}=-\dfrac{21}{10^2}\times (-1)

\dfrac{d\theta}{dt}=0.038\ rad/s

Therefore rate of change in the angle is 0.038\ rad/s

8 0
3 years ago
What causes an object to move, change direction or change speed?
Wittaler [7]
I believe it would be an unbalanced force. Because the forces are unbalanced, one side is stronger and, therefore, the object will move.
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