An object is five focal lengths from a concave mirror.how do the object and image heights compare?

1 answer:

7 0

An object distance is presented as s = 5f and we know that the mirror equation relates the image distance to the object distance and the focal length.

The mirror equation is 1/f = 1/s + 1/s’ where the variable f stands for the focal length of the mirror. Variable (s) represents the distance between the mirror surface and the object and the variable <span>(s’) represents the distance between the mirror surface and the image. </span>

In addition, a concave mirror will have a positive focal length (f) and a convex mirror will have a negative focal length (f).

Now, we then have 1/f = 1/5f + 1/s’ which is s’ = 5f/4

Then we get the magnification ratio that expresses the size or amount of magnification or reduction of the object or image and to get the magnification, we use this equation: M= s’/s

M= 5f/4x5f

s’ = 1/4s

Therefore, the image height is one fourth of the object height

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A speeding car traveling at 24.8 m/s passes a police car that is at rest. The police car begins its pursuit at the instant the s

marusya05 [52]

Answer:

the acceleration required is 1.37m/s^2

Explanation:

The car is having a constant velocity movement, so if we calculate the time to reach 897m, we can use it to find the acceleration the policeman need to apply to reach the car.

$x=v*t\\t=\frac{x}{v}\\t=\frac{897m}{24.8m/s}\\t=36.17s$

the policeman is traveling with a constant acceleration starting from rest so:

$x=\frac{1}{2}*a*t^2\\\\a=\frac{2*x}{t^2}\\\\a=1.37 m/s2$

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3 years ago

Master of physics needed

Delicious77 [7]

Hey JayDilla, I get 1/3. Here's how:
Kinetic energy due to linear motion is:
$E_{linear}= \frac{1}{2}mv^2$
where
$v=r \omega$
giving
$E_{linear}= \frac{1}{2}mr^2 \omega ^2$

The rotational part requires the moment of inertia of a solid cylinder
$I_{cyl} = \frac{1}{2}mr^2$
Then the rotational kinetic energy is
$E_{rot}= \frac{1}{2}I \omega ^2= \frac{1}{4}mr^2 \omega ^2$
Adding the two types of energy and factoring out common terms gives
$\frac{1}{2}mr^2 \omega ^2(1+ \frac{1}{2})$
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part. Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.

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3 years ago

Two dogs are pulling on a chew toy. One dog pulls the chew toy with 64 N [E] and

PIT_PIT [208]

Answer:

Eastward, at 11 m/s^2

Explanation:

64N-31N=unbalanced force of 33N

F=ma

33N=(3kg)a

a=11m/s^2 to the East

3 0