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zheka24 [161]
3 years ago
12

An 8.0 cm object is 40.0 cm from a concave mirror that has a focal length of 10.0 cm. Its image is 16.0 cm in front of the mirro

r.
Physics
1 answer:
svet-max [94.6K]3 years ago
6 0
 We can rearrange the mirror equation before plugging our values in. 
1/p = 1/f - 1/q. 
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p  <-- cross multiplication
13.33cm = p

Now that we have the value of p, we can plug it into the magnification equation.

M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'

So the height of the image produced by the mirror is 9.6cm.
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The density of a substance is 3.4 g cm-3. Its relative density relative to another substance is 2.0. what is the density of the
ankoles [38]

Answer:

1.7 g/cm³

Explanation:

Given that:

Density of substance = 3.4 g/cm³

Relative density to another substance = 2

Density of second substance=?

Let density of second substance = x

Relative density = density of substance / density of second substance

Relative density = density of substance / x

2.0 = 3.4g/cm³ / x

2 * x = 3.4 g /cm³

x = 3.4 g/cm³ ÷ 2

x = 1.7 g/cm³

5 0
3 years ago
Calculate the wavelength of light emitted when each of the following n=5 n=3
denis-greek [22]
When it transmits from n=5 to n=3, 
∧ = Rh [ni - nf]
∧ = Rh [1/3-1/5]
∧ = Rh 2/15

So, your final answer is Rh * 2/15
Where, Rh = Rydberg's constant

Hope this helps!
4 0
3 years ago
A toy car is placed 29.0cm from a convex mirror. The image of the car is upright and one-sixth as large as the actual car. Calcu
borishaifa [10]

Answer:

The power of the mirror is -17.24 diopter.

Explanation:

Magnification of an image,m = \frac{1}{6}

Distance of the object from the mirror = u = 29.0 cm

Distance of an image from the mirror = v

Magnification = m=\frac{-v}{u}

\frac{1}{6}=\frac{-v}{u}

v=\frac{-u}{6}

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{u}+\frac{-6}{u}

f=\frac{-u}{5}=\frac{-29.0 cm}{5}=-5.8 cm

Focal length of mirror = -5.8 cm =-0.058 m

Power of the lens = P

P =\frac{1}{f}

P=\frac{1}{-0.058 m}=-17.24 D

8 0
3 years ago
A car moving in a straight line starts at x = 0 at t = 0 . It passes the point x = 30.0 m with a speed of 10.0 m/s at t = 3.00 s
Bogdan [553]

Answer:

a) v=20.3m/s

b) a=2.35m/s^2

Explanation:

From the exercise we know:

x_{1}=30m; v_{1}=10m/s; t_{1}=3s

x_{2}=375m; v_{2}=50m/s; t_{2}=20s

The formula for average velocity is:

v=\frac{x_{2}-x_{1}}{t_{2}-t_{1}}

a) v=\frac{375m-30m}{20s-3s}=20.3m/s

The formula for average acceleration is:

a=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}

b) a=\frac{(50-10)m/s}{(20-3)s}=2.35m/s^2

4 0
3 years ago
Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of
WINSTONCH [101]

a₀).  You know ...
         -- the object is dropped from 5 meters
             above the pavement;
         --  it falls for 0.83 second.

a₁).  Without being told, you assume ...
         -- there is no air anyplace where the marshmallow travels,
             so it free-falls, with no air resistance;
         -- the event is happening on Earth,
            where the acceleration of gravity is  9.81 m/s² .

b).  You need to find how much LESS than 5 meters
       the marshmallow falls in 0.83 second.
    
c).  You can use whatever equations you like.
       I'm going to use the equation for the distance an object falls in
       ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d).  To see how this all goes together for the solution, keep reading:


The distance that an object falls in ' T ' seconds
when it's dropped from rest is

                                 (1/2 G) x (T²) .

On Earth, ' G ' is roughly  9.81 m/s², so in 0.83 seconds,
such an object would fall

                               (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

                         (5.00 - 3.38)  =  1.62 meters

above the pavement.  That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.

7 0
3 years ago
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