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kipiarov [429]
2 years ago
11

An object is 10. cm from the mirror, its height is 3.0 cm, and the focal length is 2.0 cm. What is the distance from the

Physics
1 answer:
Rashid [163]2 years ago
4 0

The distance from the image to the mirror is C) 2.5 cm

Explanation:

A mirror is an object that reflects the light coming from an object, creating an image of the object itself.

We can solve this problem by using the mirror equation, which states that:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where:

f is the focal length of the mirror

p is the distance between the object and the mirror

q is the distance between the image and the mirror

For the situation in  this problem, we have:

f = 2.0 cm is the focal length of the mirror

p = 10.0 cm is the distance of the object from the mirror

Solving for q, we find:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{2.0}-\frac{1}{10}=\frac{4}{10}\\\rightarrow q=\frac{10}{4}=2.5 cm

Learn more about reflection and other wave phenomena:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

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As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil
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Answer:

Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

F_1 = 5.7\ N\\\\F_2 = 1.9\ N

We have 2 unknowns (F_3 and b) and we have 2 equations.

Now we clear F_3 from the second equation and introduce it into the first equation.

F_3 = \frac{F_2}{cos (b)}

Then

-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

Then we find the value of F_3

F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N

Finally the answer is 6.3 N at 162° counterclockwise from  

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7 0
3 years ago
A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywh
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Answer:

(a) 8.362 rad/sec

(b) 6.815 m/sec

(c) 9.446 rad/sec^2

(d) 396.22 revolution

Explanation:

We have given that diameter d = 1.63 m

So radius r=\frac{d}{2}=\frac{1.63}{2}=0.815m

Angular speed N = 79.9 rev/min

(a) We know that angular speed in radian per sec

\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 79.9}{60}=8.362rad/sec

(b) We know that linear speed is given by

v=r\omega =0.815\times 8.362=6.815m/sec

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