All categories

3 years ago

A mirror forms an erect image 40cm from the object and one third its height where must the mirror be situated

Physics

1 answer:

iragen [17]3 years ago

5 0

The mirror is located 30 cm from the object

Explanation:

To solve the problem, we can use the magnification equation, which states that:

$M=\frac{y'}{y}=-\frac{q}{p}$

where

M is the magnification

y' is the size of the image

y 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 notice that the image formed by the mirror is erect and diminished: this means that this is a convex mirror, so the image is virtual, and this means that the image and the object are located on opposite sides of the mirror. Therefore,

$p-q=40 cm$ (distance of the image from the object is 40 cm, but since the image is virtual, q is the negative)

The size of the image is 1/3 that of the object, so

$y'=\frac{1}{3}y$

Solving the equation for p, we find the distance between the object and the mirror:

$\frac{y'}{y}=-\frac{p-40}{p}\\\\p=\frac{40}{1+\frac{y'}{y}}=\frac{40}{4/3}=30 cm$

So, the mirror is 30 cm from the object.

#LearnwithBrainly

You might be interested in

Which of the following elements are most likely to have similar chemical properties?

NNADVOKAT [17]

Answer:

transition metal, and inner transition metals groups are numbered 1-18 from left to right

5 0

3 years ago

After school, several students skateboard at the nearby park. Your best friend, Erwin, is a good skateboarder and happens to fal

Alchen [17]

Answer:

A person who never gives up.

Explanation:

due to his passion for skateboarding he try's his never gives up until he will finally learns the trick.

3 0

3 years ago

A chemist identifies compounds by identifying bright lines in their spectra. She does so by heating the compounds until they glo

padilas [110]

Answer:

a) λ = 189.43 10⁻⁹ m b) λ = 269.19 10⁻⁹ m

Explanation:

The diffraction network is described by the expression

d sin θ= m λ

Where m corresponds to the diffraction order

Let's use trigonometry to find the breast

tan θ = y / L

The diffraction spectrum is measured at very small angles, therefore

tan θ = sin θ / cos θ = sin θ

We replace

d y / L = m λ

Let's place in the first order m = 1

Let's look for the separation of the lines (d)

d = λ L / y

d = 501 10⁻⁹ 9.95 10⁻² / 15 10⁻²

d = 332.33 10⁻⁹ m

Now we can look for the wavelength of the other line

λ = d y / L

λ = 332.33 10⁻⁹ 8.55 10⁻²/15 10⁻²

λ = 189.43 10⁻⁹ m

Part B

The compound wavelength B

λ = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²

λ = 269.19 10⁻⁹ m

4 0

3 years ago

Can someone explain how they got their answer or how I get the change in number? :(

enyata [817]

Answer:

See Below

Explanation:

Okay, I thinkkk what it is asking by what you summarzied for me issss:

They split the total time into four quarters. They then took (for the first quarter) the start time. Then when the first quarter ends and the second quarter starts is the "end" time.

They then subtract the start time of the second quarter from the end time of the first quarter.

I hope this helps, good luck! :D

5 0

2 years ago

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 49.0 cm . The explorer finds that

iren2701 [21]

Answer: g = 10.0 m/s/s

Explanation:

For a simple pendulum, provided that the angle between the lowest and highest point of his trajectory be small, the oscillation period is given by the following expression:

T = 2π √L/g , where L = pendulum length, g= accelleration of gravity.

We can also define the period, as the time needed to complete a full swing, so from the measured values, we can conclude the following :

T = 140 sec/ 101 cycles = 1.39 sec

Equating both definitions for T, we can solve for g, as follows:

g = 4 π² L / T² = 4π². 0.49 m / (1.39)² = 10.0 m/s/s

7 0

3 years ago

A mirror forms an erect image 40cm from the object and one third its height where must the mirror be situated ​

A mirror forms an erect image 40cm from the object and one third its height where must the mirror be situated