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Cloud [144]
3 years ago
5

What an object is placed 8 mm from a concave spherical mirror a clear image can be projected on the screen 16 mm in front of me

to add you to the height of 4 mm the height of the image is
Physics
1 answer:
alexgriva [62]3 years ago
4 0

Concept: The magnification of spherical mirror can be defined by two ways.

(i) In terms of the height of the object and image.

The magnification of the spherical mirror is defined as the ratio of the height of the image'h_{i}' to the height of the object 'h_{o}'. It is denoted by letter 'm'.

Mathematically, it can be written as

m= \frac{h_{i}}{h_{o}}   ------------(1)

(ii) In terms of the object's and image's distances.

The magnification of the spherical mirror is defined as the negative ratio of the image distance'd_{i}' to the object distance 'd_{o}'.

Mathematically, it can be written as

m= - \frac{d_{i}}{d_{o}}   ------------(2)

Now, from equation (1) and (2) we have,

m = \frac{h_{i}}{h_{o}}   = -  \frac{d_{i}}{d_{o}}  -----------(3)

Given: Spherical Concave Mirror,

We will consider positive sign for object's and image's distance because both are in front of the mirror.

Object distance (d_{o}) = + 8 mm.

Image distance (d_{i}) = + 16 mm

Object's height (h_{o}) = + 4 mm

Image's height (h_{i}) =?

Now, apply equation (3)

\frac{h_{i}}{h_{o}}   = - \frac{d_{i}}{d_{o}}

Or,   \frac{h_{i}}{4 mm}   = - \frac{+16 mm}{+8 mm}

Or, hi = - 8 mm

Here; negative sign means, the image will be inverted.

The image's height will be 8 mm.

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Two children ride side-by-side on a carousel. Their paths are shown in the image below.
Sonbull [250]

Answer:

The child represented by a star on the outside path.

Explanation:

5 0
3 years ago
Which organelle is the powerhouse of the cell, the site of cellular respiration? A) 2 - nucleus B) 5 - endoplasmic reticulum C)
Paladinen [302]
D) 9- mitochondria
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7 0
3 years ago
You have a stopped pipe of adjustable length close to a taut 62.0 cmcm, 7.25 gg wire under a tension of 4710 NN. You want to adj
Oksana_A [137]

Answer:

6 cm long

Explanation:

F = 4110N

Vo(speed of sound) = 344m/s

Mass = 7.25g = 0.00725kg

L = 62.0cm = 0.62m

Speed of a wave in string is

V = √(F / μ)

V = speed of the wave

F = force of tension acting on the string

μ = mass per unit density

F(n) = n (v / 2L)

L = string length

μ = mass / length

μ = 0.00725 / 0.62

μ = 0.0116 ≅ 0.0117kg/m

V = √(F / μ)

V = √(4110 / 0.0117)

v = 592.69m/s

Second overtone n = 3 since it's the third harmonic

F(n) = n * (v / 2L)

F₃ = 3 * [592.69 / (2 * 0.62)

F₃ = 1778.07 / 1.24 = 1433.927Hz

The frequency for standing wave in a stopped pipe

f = n (v / 4L)

Since it's the first fundamental, n = 1

1433.93 = 344 / 4L

4L = 344 / 1433.93

4L = 0.2399

L = 0.0599

L = 0.06cm

L = 6cm

The pipe should be 6 cm long

6 0
3 years ago
GRAPH INCLUDED PLEASE HELP
swat32

Answer:

10 days

Explanation:

The half-life of a radioactive sample is the time taken for half of the sample to decay.

In the diagram, the half-life corresponds to the time after which the % of cobalt-57 has halved. We can observe the following:

At t=10 days, the % of Co remaining is approximately 45%

At t=20 days, the % of Co remaining is approximately 22%

This means that the sample of cobalt-57 has halved in 10 days, so the half-life of cobalt-57 is 10 days.

5 0
3 years ago
A light beam is traveling through an unknown substance. When it strikes a boundary between that substance and the air (nair≈1),
SashulF [63]

Answer:

0.7707

Explanation:

From Snell's law,

n(1) * sin θ1 = n(2) * sinθ2

Where n(1) = refractive index of air = 1.0003

θ1 = angle of incidence

n(2) = refractive index of second substance

θ2 = angle of refraction

The angle of reflection through the unknown substance is the same as the angle of incidence of air. Thus this means that θ1 = 29°

=> 1.0003 * sin29 = n(2) * sin39

n(2) = (1.0003 * sin29) / sin39

n(2) = 0.7707

Explanation:

6 0
3 years ago
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