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ankoles [38]
4 years ago
12

Write the word AMBULANCE as it would appear when reflected in a plane mirror

Physics
1 answer:
Lady_Fox [76]4 years ago
7 0

ECNALUBMA

Explanation:

When the word AMBULANCE is reflected on a plane mirror, it is shown as ECNALUBMA.

 On a plane mirror, lateral inversion occurs and the position of the words are interchanged.

 This is why it is written as ECNALUBMA on the front of  Ambulance buses.

When scooters and drivers check their side mirror, they will see the properly written ambulance word due to lateral inversion.

Learn more:

Mirror adjustment brainly.com/question/8737441

#learnwithBrainly

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A loudspeaker has a circular opening with a radius of 0.1158 m. The electrical power needed to operate the speaker is 27.0 W. Th
Vilka [71]

Answer:

electrical power is converted by the speaker into sound power 17.25 %

Explanation:

given data

radius = 0.1158 m

power needed = 27.0 W

average sound intensity = 25.6 W/m²

solution

we know sound intensity that is

I = \frac{power}{area}   ...................1

power = intensity × area

Power = 25.6 × 4 ×π×r²

Power = 25.6 × 4 ×π×0.1158²

power = 4.313 W

so here electric power % into sound power

= \frac{4.313}{25} × 100

= 17.25 %

3 0
3 years ago
A wire is stretched 30% what is the percent age change in resistance?​
IgorC [24]

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3 years ago
Rank the deformations of the following rods in terms of the magnitude of the average normal strain: (a) The length of a 1-m-long
kipiarov [429]

To solve this problem we will consider the concepts related to the normal deformation on a surface, generated when the change in length is taken per unit of established length, that is, the division between the longitudinal fraction gained or lost, over the initial length. In general mode this normal deformation can be defined as

\epsilon = \frac{\delta}{l} = \frac{l_0-l}{l}

Here,

\delta= Change in final length (l_0) and the initial length l

PART A)

\epsilon = \frac{\delta_1}{l}

\epsilon = \frac{l_0-l}{l_0}

\epsilon = \frac{1.02-1}{1}

\epsilon = 0.01961

PART B)

\epsilon = \frac{\delta_1}{l}

\epsilon = \frac{l_0-l}{l_0}

\epsilon = \frac{2-1.05}{2}

\epsilon = 0.475

PART C)

\epsilon = \frac{\delta_1}{l}

\epsilon = \frac{l_0-l}{l_0}

\epsilon = \frac{3.07-3}{3}

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Therefore the rank of this deformation would be  B>C>A

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3 years ago
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Answer: The correct answer is B- The ions are free to move in water and carry an electric charge.

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