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3 years ago

Which statement is true about the part of the electromagnetic spectrum that human eyes can detect?

Physics

2 answers:

LUCKY_DIMON [66]3 years ago

7 0

Your answer is B. The human eye can only detect color and Tv waves

Mama L [17]3 years ago

4 0

Answer:

The answer is D i took the quiz and I am in K12 hope it helps!

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What is dispersion of light?

algol13

Answer:

$\huge \bold \blue{ \underline{ answer}}$

The splitting up of light into its constituent colours while passing from one medium to the other is called dispersion.

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2 years ago

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A liquid at rest in a fixed container exerts a force perpendicular to the wall of the container. Two students make claims about

Svetllana [295]

Answer:

Explanation:

Given that a liquid at rest in a fixed container exerts a force perpendicular to the wall of the container. Two students make claims about the microscopic cause of this force. Student A says that the force exerted by individual molecules as they bounce off the wall is always perpendicular to the wall. Student B says that the molecules may strike the wall at angles that are not perpendicular. Which student is correct and why?

The correct answer is option C

Student B . The molecules in a liquid are in random motion at the microscopic scale. For every atom that hits at an angle to one side of perpendicular, there is likely to be another atom hitting at the same speed at the same angle on the other side of perpendicular. On average the nonperpendicular components cancel

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4 years ago

When the energy of a sound wave is transferred to a particle of a medium and causes it to vibrate and generate heat the sound wa

skelet666 [1.2K]

The answer of this said question is d

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3 years ago

What voltage produces a current of 50 amps with a resistance of 20 ohms?<br>

PIT_PIT [208]

Answer:

1000 V

Explanation:

Ohm's law:

V = IR

V = (50 A) (20 Ω)

V = 1000 V

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3 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

4 years ago