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Which of the following is not a reason fluorescent lamps are advantageous over incandescent lamps?

cricket20 [7]

Answer;

B. Fluorescent lamps operate at a higher temperature than incandescent lamps.

Explanation;

-A fluorescent lamp, is a type of electric light (lamp) that uses ultraviolet emitted by mercury vapor to excite a phosphor, which emits visible light.

-A fluorescent lamp produces less heat, thus, it is much more efficient. A fluorescent bulb can produce between 50 and 100 lumens per watt. This makes fluorescent bulbs four to six times more efficient than incandescent bulbs.

-Fluorescent lamps operate best around room temperature. At much lower or higher temperatures, efficacy decreases.

4 0

2 years ago

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A wave that can travel with or without a medium is called a(n)

denis23 [38]

Answer:

C

Explanation:

Answer A: A surface wave is a wave that travels along the surface of a medium.

Answer B, C: Electromagnetic waves are waves that have no medium to travel whereas mechanical waves need a medium for its transmission.

Answer D: The sentence in the answer D does not fit to the blank in the definition ( of the question )

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Hope this answer can help you.

5 0

3 years ago

The handles of the tools like screwdrivers and pliers used by electricians for repair work usually have plastic or rubber covers

fenix001 [56]

Answer:

Rubber or plastic covers are bad conductors of electricity. So they do not allow the electric current to pass through it.

Explanation:

Rubber and plastic are bad conductors of electricity, therefore when handling a tool with a rubber handle, the electricity will not pass through it.

4 0

2 years ago

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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

3 years ago

How could you make a sound wave sound louder

kykrilka [37]

If you clap your hands, the shock causes the air around your hands to begin vibrating. When air particles vibrate, they bump into other particles near them. Then these particles begin to vibrate and bump into even more air particles. When the air particles begin vibrating the air inside your ear, you hear a sound.

6 0

2 years ago