Ask question

Ask question

All categories

3 years ago

13

The virtual image produced by a convex mirror is one third the size of the object.(a) If the object is 51 cm in front of the mir

ror, what is the image distance? (Include the sign of each answer.)cm(b) What is the focal length of this mirror?cm

1 answer:

Nimfa-mama [501]3 years ago

4 0

(a) -17 cm

Magnification equation:

$\frac{h_i}{h_o}=-\frac{q}{p}$

where

$h_i$ is the size of the image

$h_o$ is the size of the object

$q$ is the distance of the image from the mirror

$p$ is the distance of the object from the mirror

In this problem, we have:

p = 51 cm (the object is 51 cm in front of the mirror)

$\frac{h_i}{h_o}=\frac{1}{3}$

So, we can find q, the image distance:

$q=-\frac{h_i}{h_o}p=-\frac{1}{3}(51 cm)=-17 cm$

and the negative sign means the image is virtual.

(b) -25.5 cm

Mirror equation:

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

where

p = 51 cm

q = -17 cm

Substituting and re-arranging the equation, we find the focal length of the mirror:

$\frac{1}{f}=\frac{1}{51 cm}-\frac{1}{17 cm}=-\frac{2}{51 cm}\\f=-\frac{51 cm}{2}=-25.5 cm$

and the negative sign is due to the fact it is a convex mirror.

You might be interested in

A certain organ pipe, open at both ends, produces a fundamental frequency of 288 Hz in air. Part A If the pipe is filled with he

alina1380 [7]

Answer:

773.25 Hz

Explanation:

Concept : In an open organ pipe in fundamental mode of vibration

wave length of wave λ = 2L

where L is length of the pipe

frequency = velocity of sound / λ

Given values: fundamental frequency = 288 Hz

fluid is air. velocity of sound = 340 m/s

⇒ 288 = 340/2L

⇒L = 59.02 cm

The point to be noted is if the pipe is filled with helium initially at the same temperature, there would be change in the sound velocity .Then, frequency of note produced will also be changed .

We know that velocity of sound is inversely proportional to square root of molar mass of gas

velocity of sound in air / velocity of sound in helium = Square root of (Molar mass of Helium/ molar mass of air)

$\frac{V_a}{V_{He}} = \sqrt{\frac{4}{28.8} } \\\frac{340}{V_{He}} =0.3725\\V_{He} =912.5 m/s$

Now, frequency = velocity of sound / λ

= 912.75 / (2 x 0.5902)

= 773.25 Hz

6 0

3 years ago

The strength of an electric field depends on the

KengaRu [80]

<span>The correct option is D) both A and B. In fact, the strength of an electric field produced by a single point charge is given by
</span> $E=k \frac{Q}{r^2}$ <span>
where Q is the charge that generates the field, k is the Coulomb's constante and r is the distance from the charge source of the field.

From the equation, we see that the strength of the field depends on A) the amount of charge that produced the field (Q) and B) the distance from the charge (r), so the correct option is D.</span>

5 0

3 years ago

Read 2 more answers

The figure below shows a combination of capacitors. Find (a) the equivalent capacitance of combination, and (b) the energy store

velikii [3]

Answer:

A) C_{eq} = 15 10⁻⁶ F, B) U₃ = 3 J, U₄ = 0.5 J

Explanation:

In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.

In this case let's start by finding the equivalent capacitance.

A) Let's solve the part where C1 and C3 are. These two capacitors are in serious

$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}$ (you has an mistake in the formula)

$\frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \ 10^{6}$

$\frac{1}{C_{eq1}}$ = 0.1 10⁶

$C_{eq1}$ = 10 10⁻⁶ F

capacitors C₂, C₄ and C₅ are in series

$\frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}$

$\frac{1}{C_{eq2} } = (\frac{1}{15} + \frac{1}{30} + \frac{1}{10} ) \ 10^6$

$\frac{1}{C_{eq2} }$ = 0.2 10⁶

$C_{eq2}$ = 5 10⁻⁶ F

the two equivalent capacitors are in parallel therefore

C_{eq} = C_{eq1} + C_{eq2}

C_{eq} = (10 + 5) 10⁻⁶

C_{eq} = 15 10⁻⁶ F

B) the energy stored in C₃

The charge on the parallel voltage is constant

is the sum of the charge on each branch

Q = C_{eq} V

Q = 15 10⁻⁶ 6

Q = 90 10⁻⁶ C

the charge on each branch is

Q₁ = Ceq1 V

Q₁ = 10 10⁻⁶ 6

Q₁ = 60 10⁻⁶ C

Q₂ = C_{eq2} V

Q₂ = 5 10⁻⁶ 6

Q₂ = 30 10⁻⁶ C

now let's analyze the load on each branch

Branch C₁ and C₃

In series combination the charge is constant Q = Q₁ = Q₃

U₃ = $\frac{Q^2}{2 C_3}$

U₃ = $\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}$

U₃ = 3 J

In Branch C₂, C₄, C₅

since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅

U₄ = $\frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}$

U₄ = 0.5 J

7 0

3 years ago

Essay about why people should not join a gang 300 word

Salsk061 [2.6K]

Answer:

I don't know if it's fair for me to write an entire essay for you

but if you would like I can list some reasons you can incorporate into an essay.

-------------------------------------------------------------------------------------------------------

-If you are joining a gang you are most likely going to have a greater chance of being targeted, even if it seems like you would be protected.

-Most of the time when someone joins a gang a child is being separated from a parent and will face most aggression.

-This can also cause academic failure in students, from lack of education and probably a pretty bad background.

-For teens when joining a gang they will do things that aren't suited for their age (I would list things but I'm not trying to get myself reported if you know what I mean)

-You will also have a great risk of being imprisoned and facing bad consequences, and these consequences can even be having the death penalty.

-------------------------------------------------------------------------------------

I hope this is good enough, sorry I didn't give you a full essay. I think this should help with that though.

6 0

3 years ago

Young's double slit experiment is one of the quintessential experiments in physics. The availability of low cost lasers in recen

ddd [48]

Answer:

Fringe width = 21 mm

Explanation:

Fringe width is given by the formula

$\beta = \frac{\Lambda L}{d}$

here we know that

$\Lambda = 564 nm$

L = 4.0 m

d = 0.108 mm

now from above formula we will have

$\beta = \frac{(564 \times 10^{-9})(4.0 m)}{0.108\times 10^{-3}}$

$\beta = 0.021 meter$

so fringe width on the wall will be 21 mm

7 0

3 years ago