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astraxan [27]
4 years ago
15

A statue is made of bronze, which is an alloy of about 80% copper and 20% tin. Which metal is in the higher concentration in thi

s bronze statue?
A.The concentrations of each metal in the statue are equal.
B.The statue has a higher concentration of copper.
C.Tin is more concentrated in the statue.
Physics
2 answers:
Olenka [21]4 years ago
7 0

Answer:

<h2>B. The statue has a higher concentration of copper.</h2>

Explanation:

We know by given that the statue is made of bronze, which is formed by 80% copper and 20% tin.

So, as you can observe, the copper represents a majority of the statue, that means it has more concentration of copper than tin.

Therefore, the right answer is B. The statue has a higher concentration of copper, because it's made 80% of copper actually.

skad [1K]4 years ago
3 0
The answer is B because it's 80 percent copper, that's a greater number then 20.
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Answer:

<em>the phase relationship between two waves.</em>

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Coherence describes all properties of the correlation between physical quantities between waves. It is an ideal property of waves that determines their interference. In a situation in which there is a correlation or phase relationship between two waves. If the properties of one of the waves can be measure directly, then, some of the properties of the other wave can be calculated.

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3 years ago
An electron in the n = 5 level of an h atom emits a photon of wavelength 1282.17 nm. to what energy level does the electron move
lions [1.4K]
This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
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Some rearranging and collecting up terms:
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8 0
3 years ago
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F=ma

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Number 2

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4 0
3 years ago
Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B i
mel-nik [20]

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2}  =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}

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w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}

The velocity at the contact point is equal to:

v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}

v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}

Matching both expressions:

1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s

b) The time during which the disks slip is:

t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s

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\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)

6 0
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