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3 years ago

which type of wave spreading do you think causes faster energy loss-two-dimensional or three-dimensional? explain.

Physics

2 answers:

madreJ [45]3 years ago

4 0

Answer:

Three dimensional wave

Explanation:

We know that waves generally spread in multiple dimensions and have different wavefronts, the more a wavefront moves away from the source it spreads the energy as it becomes larger hence it starts loosing energy.

As the number of wavefront in a three dimensional wave will certainly be more than that of two dimensional wave, it will spread more energy and will definitely starts to loose its energy way faster than two dimensional wave

sdas [7]3 years ago

3 0

Three dimensional would loose faster

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A ball is gently dropped from a height of 20 m. If its velocity Increases uniformly at the rate of 10 m s2, with what velocity w

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Answer:

20 m/s

Explanation:

Height (s)= 20m

acceleration (a) = 10 m/s^2

v=?

we know,

V^2= U^2 + 2as. (U is initial velocity)

or,. V^2 = 0 + 2 × 10 × 20

or, V^2 = 400

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2 years ago

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Make the following conversion

ValentinkaMS [17]

Answer:

The value of 60.2 mg in to g is 0.0602.

Explanation:

We need to calculate the 60.2 mg in to gram.

Conversion mg to g :

1 g = 1000 mg

Using this conversion,

$1 mg = 10^{-3}g$

So, $60.2\ mg = 60.2\times10^{-3}\ g$

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Hence, The value of 60.2 mg in to g is 0.0602.

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3 years ago

Name and draw the circuit symbols for the

pishuonlain [190]

Answer:

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Explanation:

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3 years ago

Determine the values of m and n when the following average distance from the Sun to the Earth is written in scientific notation:

Leviafan [203]

Answer:

$150000000000\ m=1.5\times 10^{11}\ m$

Explanation:

A number can be written in the form of :

$N=m\times 10^n$

Where

m is the real number

n is any integer

In this case, the average distance from the Sun to the Earth is given,

d = 150000000000 meters

There are 10 zeroes in this number. We need to write this number in scientific notation. It is given by :

$d=1.5\times 10^{11}\ m$

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3 years ago

A whale swims due east for a distance of 6.9 km, turns around and goes due west for 1.8 km, and finally turns around again and h

xeze [42]

The question is incomplete. Here is the complete question:

A whale swims due east for a distance of 6.9 km, turns around and goes due west for 1.8 km and finally turns around again and heads 3.7 km due east. (a) What is the total distance traveled by the whale? (b) What are the magnitude and direction of the displacement of the whale?

Answer:

(a) Distance = 12.4 km

(b) Displacement = 8.8 km due east

Explanation:

Consider east direction as positive and west direction as negative.

Given:

The motion is along the east-west line.

The whale first swims 6.9 km due east, then 1.8 km due west and again 3.7 km due east.

(a)

Distance traveled by the whale is equal to the sum of the lengths of all the distances traveled. Therefore,

Distance traveled = 6.9 km + 1.8 km + 3.7 km = 12.4 km

Therefore, the distance traveled by the whale is 12.4 km.

(b)

Displacement of the whale is given by considering the sign of each of the individual displacements. Therefore,

Displacement of the whale = (+6.9 km) + (-1.8 km) + (+3.7 km)

Displacement of the whale = 6.9 km - 1.8 km + 3.7 km = 8.8 km

The answer is positive. So, the direction is due east.

Therefore, the displacement of the whale is <u>8.8 km due east.</u>

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3 years ago