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Valentin [98]
3 years ago
9

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

Physics
1 answer:
enot [183]3 years ago
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

7 0
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.

8 0
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
2 years ago
The minimum stopping distance of a car moving at 20.5 mi/h is 11.6 m. Under the same conditions (so that the maximum braking for
pshichka [43]

Answer:

d = 69 .57 meter

Explanation:

First case

Speed of car ( v )  = 20.5 mi/h  = 9.164  M/S

distance ( d ) = 11.6 meter                                       ( m = mass of the car )

Work done = 0.5 m v²  = 0.5 * 9.164² * m J  = 41.99 m J

Force = ( workdone /distance ) = ( 41.99 m / 11.6 )   =  3.619 m N

Second case

v = 50.2 mi/h = 22.44135 m/s

d = ?

Work done = 0.5 * 22.44² * m J = 251.7768 * m J

Since the braking force remains the same .

3.619 m = ( 251.7768 m / d )

d = 69 .57 meter

7 0
3 years ago
An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elect
beks73 [17]

Answer:

Speed of the alpha particle is v=1.8180\times 10^3m/sec      

Explanation:

We have given charge on alpha particle q=3.2\times 10^{-19}C

Mass of the alpha particle m=6.68\times 10^{-27}kg

Potential difference V=-3.45\times 10^{-3}volt

We have to find the speed of the alpha particle

From energy conservation we know that

\frac{1}{2}mv^2=qV

\frac{1}{2}\times 6.68\times 10^{-27}\times v^2=3.2\times 10^{-19}\times 3.45\times 10^{-3}

v=1.8180\times 10^3m/sec

4 0
3 years ago
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