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Deffense [45]
2 years ago
12

The Statue of Liberty is made of copper that has turned green because it had undergone a change. What can be said about this cha

nge? A. The change is a physical change because the copper in the statue is still there. B. The change is a physical change because a new chemical is formed on the outside of the statue. C. The change is not a physical change because the color of the statue has changed. D. The change is not a physical change because a new chemical is formed in the outside of the statue.
Physics
2 answers:
Mekhanik [1.2K]2 years ago
7 0

Answer:

D

Explanation:

Sergeeva-Olga [200]2 years ago
3 0
D. because the statue has a new chemical called patina. It is no longer copper.
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3 years ago
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ASHA 777 [7]

Answer:

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Explanation:

After each half life, the mass is reduced by half.

2/2 = 1

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2 years ago
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A student wants to determine the speed of sound at an elevation of one mile. To do this the student performs an experiment to de
Lorico [155]

Answer:

The correct answer is a

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For the same medium, the speed of sound depends on the temperature of the fora

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Answer:

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Water is flowing through a 45° reducing pipe bend at a rate of 200 gpm and exits into the atmosphere (P2 = 0 psig). The inlet to
Nataliya [291]

Answer:

F1=177.88 Newtons

Explanation:

Let's start with the Bernoulli's equation:

P_{1} + \frac{1}{2}\beta V_{1} ^{2} + \beta gh_{1}  =P_{2} + \frac{1}{2}\beta V_{2} ^{2} + \beta gh_{2}

Where:

P is pressure, V is Velocity, g is gravity, h is height and β is density (for water β=1000 kg/m3); at the points 1 and 2 respectively.

From the Bernoulli's equation and assuming that h is constant and P2 is zero (from the data), we have:

P_{1} + \frac{1}{2}\beta V_{1} ^{2} = \frac{1}{2}\beta V_{2} ^{2}

As we know, P1 must be equal to \frac{F_{1} }{A_{1}}, so, replacing P1 in the equation, we have:

P_{1} = \frac{F_{1}}{A_{1}} = \frac{1}{2}\beta(V_{2} ^{2} - V_{1} ^{2})

And

F_{1} = {A_{1}} ( \frac{1}{2}\beta(V_{2} ^{2} - V_{1} ^{2}))

Now, let's find the velocity to replace the values on the expression:

We can express the flow in function of velocity and area as Q = V A, where Q is flow, V is velocity and A is area. As the same, we can write this: Q_{1} = V_{1} A_{1}\\Q_{2} = V_{2} A_{2}. In the last two equations, let's clear Velocities.

V_{1} = \frac{Q_{1}}{A_{1}}\\V_{2} = \frac{Q_{2}}{A_{2}}

and replacing V1 and V2 on the last equation resulting from Bernoulli's (the one that has the force on it):

F_{1} = {A_{1}} ( \frac{1}{2}\beta((\frac{Q_{2}}{A_{2}})^{2} - (\frac{Q_{1}}{A_{1}})^{2}))

First, we have to consider that from a mass balance, the flow is the same, so Q1=Q2, what changes, is the velocity. Knowing this, let's write the areas, diameters, density and flow on International Units System (S.I.), because the exercise is asking us the answer in Newtons.

D_{1}=1.5 inches=0.0381 mts\\D_{2}=1 inches=0.0254 mts\\A_{1}=\frac{\pi D_{1}^{2} }{4}=0.00114mts^{2}\\A_{2}=\frac{\pi D_{2}^{2} }{4}=0.000507mts^{2}\\\beta=1000 kgs/m^{3}\\Q=200gpm=0.01mts^{3}/seg

Replacing the respective values in this last expression, we obtain:

F1 = 177.88 N

3 0
3 years ago
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