Option C would be the right one
The density of the gold is calculated to be "19,300 kg/m³".
<u>Explanation:</u>
Given:
Volume = 2cm³
Mass = 38.6 grams.
To Find:
Density of the gold = ?
Solution:
Density is obtained by dividing mass of the sample by its volume and it is given in the units of kg/m³.
Mass in grams is converted into kg as,
1 g = 0.001 kg
38. 6 g = 
Now we have to convert cm³ to m³ as,
1 cm³ = 10⁻⁶ m³
2 cm³ = 2 × 10⁻⁶ m³
So Density = 
Physical properties of Gold:
- It is Malleable and ductile.
- It is a corrosion resistant element.
Answer:
Electric potential energy
Explanation:
The Electric potential energy of a system is less than that carried out by electrostatic forces during the development of the system (as long as charges are initially cut infinitely).
The change in potential energy between an initial and final configuration is equal to minus the work done by the electrostatic forces.
Answer:
1.16cm were cut off the end of the second pipe
Explanation:
The fundamental frequency in the first pipe is,
<em><u>Since the speed of sound is not given in the question, we would assume it to be 340m/s</u></em>
f1 = v/4L, where v is the speed of sound and L is the length of the pipe
266 = 340/4L
L = 0.31954 m = 0.32 m
It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.
<u>So, the length of the second pipe is L – L’</u>
Then, the fundamental frequency in the second pipe is
f2 = v/4(L - L’)
<u>The beat frequency due to the fundamental frequencies of the first and second pipe is</u>
f2 – f1 = 10hz
[v/4(L - L’)] – 266 = 10
[v/4(L – L’)] = 10 + 266
[v/4(L – L’)] = 276
(L - L’) = v/(4 x 276)
(L – L’) = 340/(4 x 276)
(L – L’) = 0.30797
L’ = 0.31954 – 0.30797
L’ = 0.01157 m = 1.157 cm ≅ 1.16cm
Hence, 1.16 cm were cut from the end of the second pipe
