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azamat
3 years ago
12

What is cocanve mirror?​

Physics
2 answers:
pochemuha3 years ago
8 0

Answer:

A mirror that has a reflecting surface that is recessed inward is called concave mirror

professor190 [17]3 years ago
4 0

ehh ci menu jo kita ci answer

and oh dusra a ohh o's ne answer kita ci prr o's ton baad hor vi answer kite h.oe ne osne

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The resistance of a given conductor depends on its electrical resistivity (\rho), its length(L) and its cross-sectional area (A), as follows:

R=\frac{\rho L}{A}

In this case, we have L'=138L, \rho'=\rho and A'=A. So, the total resistance of the wire with length of 138m is:

R'=\frac{\rho' L'}{A'}\\R'=\frac{\rho 138L}{A}\\R'=138\frac{\rho L}{A}\\R'=138R\\R'=138(0.24\Omega)\\R'=33.12\Omega

5 0
3 years ago
At what water temperature will additional heat energy need to be added before the temperature will change again?
Lostsunrise [7]
That would be 0 degrees Celsius aka the melting point of water.... If you look at the diagram I attached you notice that at 0 degrees Celsius it is flat, this is because much heat is needed at this point for water to rise to 1 degree... It is the same for the boiling point (100)<span />

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3 years ago
What is the potential energy of a 30 Newton ball that is on the ground
Juli2301 [7.4K]
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3 0
3 years ago
What are the differences and relationships between speed, velocity, and acceleration
melamori03 [73]

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3 0
3 years ago
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
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