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3 years ago

A 5.0 cm object is 6.0 cm from a convex lens, which has a focal length of 7.0 cm.

Physics

2 answers:

lorasvet [3.4K]3 years ago

6 0

The distance of the image from the lens is -42 cm

The height of the image is 35 cm

Reil [10]3 years ago

5 0

Answer : The distance of the image from the lens is, -42 cm

The height of the image is, 35 cm

Solution :

First we have to calculate the image distance.

Formula used :

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

where,

f = focal length = 7 cm

p = object distance = 6 cm

q = image distance = ?

Now put all the given values in the above formula, we get the distance of the image from the lens.

$\frac{1}{7cm}=\frac{1}{6cm}+\frac{1}{q}$

$q=-42cm$

Therefore, the distance of the image from the lens is, 42 cm and the negative sign indicates that the image is virtual.

Now we have to calculate the height of the image.

Formula used :

$\frac{h}{h'}=\frac{p}{q}$

where,

h = height of object = 5 cm

h' = height of image = ?

Now put all the given values in this formula, we get the height of the image.

$\frac{5cm}{h'}=\frac{6cm}{-42cm}=-35cm$

Therefore, the height of the image is, 35 cm and the negative sign indicates that the image is inverted.

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From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s

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Let the slope of the best fit line be represented by ' $m_{best}$ '

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$m_{worst}$ = 1.29 m/s

Then the uncertainity in the slope of the line is given by the formula:

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3 years ago

Consider the points below. P(1, 0, 1), Q(−2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the poi

kozerog [31]

Answer:

a) (0, -33, 12)

b) area of the triangle : 17.55 units of area

Explanation:

We know that the cross product of linearly independent vectors $\vec{A}$ and $\vec{B}$ gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors $\vec{A} = \vec{P}-\vec{Q}$ and $\vec{B} = \vec{R} - \vec{Q}$ are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

$\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)$

$\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)$

$\vec{A} \times \vec{B} = (A_y B_z - B_y A_z) \ \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( (-1) * 3 - 1 * (-3) ) \ \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( - 3 + 3 ) \ \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}$

$\vec{A} \times \vec{B} = 0 \ \hat{i} - 33 \ \hat{j} + 12 \ \hat{k}$

$\vec{A} \times \vec{B} =(0, -33, 12)$

We know that $\vec{A}$ and $\vec{B}$ are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

$|\vec{A} \times \vec{B} | = 2 * area_{triangle}$

so:

$\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}$

$\sqrt{1233} = 2 * area_{triangle}$

$35.114= 2 * area_{triangle}$

$17.55 \ units \ of \ area = area_{triangle}$

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an object is placed at 19.3 cm in front of a diverging lens with a focal length of -18.7 cm. what is the magnification? enter a

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The magnification <u>is 31.16.</u>

Magnification is the process of increasing the apparent size of something rather than its physical size. This increase is quantified by a calculated number, also called the "factor". If this number is less than 1, it means size reduction, sometimes called size reduction or reduction.

u = -19.3

f = -18.7 cm.

m = f/f-u

= -18.7/-18.7 +19.3

The term magnification refers to the size of the image produced by the lens compared to the size of the object. For lenses: Magnification "m" is the ratio of image height to object height. The magnification of a lens is defined as the ratio of image height to object height. It is also given by image distance and object distance. equal to the ratio of image distance to object distance.

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