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2 years ago

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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.

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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For years, the tallest tower in the United States was the Phoenix Shot Tower in Baltimore, Maryland. The shot tower was used fro

serious [3.7K]

Answer:

The answer to your question is:

a) t = 3.81 s

b) vf = 37.4 m/s

Explanation:

Data

height = 71.3 m = 234 feet

t = 0 m/s

vf = ?

vo = 0 m/s

Formula

h = vot + 1/2gt²

vf = vo + gt

Process

a)

h = vot + 1/2gt²

71.3 = 0t + 1/2(9.81)t²

2(71.3) = 9,81t²

t² = 2(71.3)/9.81

t² = 14.53

t = 3.81 s

b)

vf = 0 + (9.81)(3.81)

vf = 37.4 m/s

3 0

3 years ago

Si el periodo de oscilación de resorte es de 0,44 segundos cuando oscila atado a una masa de 2 Kg. ¿Cuál será el valor de la con

boyakko [2]

Answer:

i d k h b u lol I wish I knew it sorry

7 0

2 years ago

A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr

Molodets [167]

Answer:

Explanation:

Given

$P_1=150 kPa$

$T_1=20^{\circ}C$

$V_1=0.5 m^3$

$T_2=140^{\circ}C$

$P_2=400 kPa$

R for Helium $R=2.076$

$c_v=3.115 kJ/kg-K$

mass of gas $m=\frac{P_1V_1}{RT_1}$

$m=\frac{150\times 0.5}{2.076\times 293}$

$m=0.123 kg$

Similarly $V_2$ can be found

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=0.264 m^3$

Work done $W=\int_{V_1}^{V_2}PdV$

$W=\frac{P_2V_2-P_1V_1}{n-1}$

$W=\frac{mR(T_2_T_1)}{n-1}$

Since it is a polytropic Process

therefore $PV^n=c$

$P_1V_1^n=P_2V_2^n$

$(\frac{V_1}{V_2})^n=\frac{P_2}{P_1}$

$(\frac{0.5}{0.264})^n=\frac{400}{150}$

$n=\frac{\ln 2.66}{\ln 1.893}$

$n=1.533$

$W=\frac{0.123\times 2.076(140-20)}{1.533-1}$

$W=57.48 kJ$

From Energy balance

$E_{in}-E_{out}=\Delta E_{system}$

Neglecting kinetic and Potential Energy change

$Q_{in}+W_{in}=change\ in\ Internal\ Energy$

Change in Internal Energy $\Delta U=u_2-u_1$

$\Delta U=mc_v(T_2-T_1)$

$\Delta U=0.123\times 3.115(140-20)$

$\Delta U=45.977 kJ$

$Q_{in}+57.48=45.977$

$Q_{in}=-11.50 kJ$

i.e. Heat is being removed

3 0

3 years ago

Infrared light of wavelength 2.5 µm illuminates a 0.20-mm-diameter hole. What is the angle of the first dark fringe in radians?

stepan [7]

Answer:

Θ=0.01525 rad

or

Θ=0.87°

Explanation:

Given data

wavelength λ=2.5 µm =2.5×10⁻⁶m

Diameter d=0.20 mm =0.20×10⁻³m

To find

Angle Θ in radians and degree

Solution

Circular apertures have first dark fringe at

Θ=(1.22λ)/d

Substitute the given values

So

Θ=[1.22(2.5×10⁻⁶m)]/0.20×10⁻³m

Θ=0.01525 rad

or

Θ=0.87°

6 0

3 years ago

Case 1: A 0.780-kg silver pellet with a temperature of 85 oC is added to 0.150 kg of water in a copper cup of unknown mass. The

suter [353]

Answer: The silver pellet will release heat

Explanation:

Based on the case scenario, the silver pellet has a higher temperature that the system of water and copper cup and is thereby added to the system. Because of the higher kinetic energy of the molecules of silver in the silver pellet, some of energy will be released to the water and copper cup system because the system will aim to achieve thermal equilibrium.

4 0

3 years ago