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Semenov [28]
2 years ago
15

5. Your friend claims that the Moon’s repeated orbit around Earth causes the cycle of the Moon’s phases.

Physics
2 answers:
Vaselesa [24]2 years ago
8 0

Answer:

Yes I agree

Explanation:

  • When moon rotate around earth, it appears in different phases
Anna007 [38]2 years ago
5 0

Answer:

Yes I agree with the friend, but i also disagree

Explanation:

When moon rotate around earth, it appears in different phases, because of the suns light it will show to us

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A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she
lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

f_{(1  \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}}  }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}  = 1.3225 \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz  = 740.6 \  Hz

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, L_{new} = 2/3·L

The value of the fundamental frequency will then be given as follows;

f_{(1  \, new)} =  \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \times \dfrac{2 \times L}{3} }  =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz =  840 \ Hz

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0
3 years ago
What is an effect of an object getting wet?
djverab [1.8K]

Answer:

once light hits a wet shirt, that water layer causes less of the blue shirt's blue wavelengths of light to be reflected toward your eyes and more of the blue light to be refracted, or bounce away from you, back into the fabric.

Explanation:

7 0
3 years ago
What does the path of an object look like when it is in uniform motion?​
marysya [2.9K]

Answer:

The path of an object in uniform motion is a straight line.

4 0
2 years ago
Read 2 more answers
A simple and common technique for accelerating electrons is shown in the figure, where there is a uniform electric field between
Sunny_sXe [5.5K]

Answer : 4.483\times 10^{15}\ m/s^2.

Explanation:

It is given that,

Electric field strength, E=2.55\times 10^{4}\ N/C

We know that,

Charge of electron, q=1.6\times 10^{-19}\ C

Mass of electron, m=9.1\times 10^{-31}\ kg

From the definition of electric field, F=qE...............(1)

According to Newton's second law, F = ma..........(2)

From equation (1) and (2)

ma=qE

a=\dfrac{qE}{m}

a=\dfrac{1.6\times 10^{-19}\ C\times 2.55\times 10^{4}\ N/C }{9.1\times 10^{-31}\ kg}

a=0.4483\times 10^{16}\ m/s^2

or

a=4.483\times 10^{15}\ m/s^2

So, the horizontal component of acceleration of an electron is 4.483\times 10^{15}\ m/s^2.

Hence, it is the required solution.

7 0
3 years ago
Help plzzzzzzz i need thissssssssss
frutty [35]

Answer:

The final graph

Explanation:

The graph that curves downwards is negative acceleration. While the position decreases the slop increases.

8 0
3 years ago
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