C
Atomic radius is the distance between the center of the nucleus to the outermost orbital shell of the atom. Assume the atom is like a football stadium and the nucleus of the atom is a ball placed at the center of the pitch. The atomic radius is from the center of the ball to the edge of the football stadium.
Explanation:
This atomic radius decreases from left to right of a periodic table because of increases in protons in the nucleus along the periodic table. This increased proton count has a higher attractive force on the electron orbitals of the atom. This decreases the atomic radius
The atomic radius of atoms down a column of the periodic table increase because an extra orbital shell is added to the atoms with every period down the column.
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Consider 20 deg.C. as room temperature.
From tables,
Silver has a resistivity of 1.6*10^-8 ohm-m at 20 deg.C, and it increases by 0.0038 ohm-m per deg.K increase.
Therefore if the temperature rise above 20 deg.C is T, then silver will have resistivity of
1.6*10^-8(1 + 0.0038T) ohm-m
At room temperature, the resistivity of tungsten (from tables) is 5.6*10^-8.
The resistivity of silver will be 4 times that of tungsten (at room temperature) when
1.6*10^-8(1 + 0.0038T) = 4*5.6*10^-8
1 + 0.0038T = 14
T = 13/.0038 = 3421 deg.K approx
Answer: 20 + 3421 = 3441 °C
Answer:
volume of the bubble just before it reaches the surface is 5.71 cm³
Explanation:
given data
depth h = 36 m
volume v2 = 1.22 cm³ = 1.22 × 
m³
temperature bottom t2 = 5.9°C = 278.9 K
temperature top t1 = 16.0°C = 289 K
to find out
what is the volume of the bubble just before it reaches the surface
solution
we know at top atmospheric pressure is about P1 = 
Pa
so pressure at bottom P2 = pressure at top + ρ×g×h
here ρ is density and h is height and g is 9.8 m/s²
so
pressure at bottom P2 = + 1000 × 9.8 ×36
pressure at bottom P2 =4.52 × Pa
so from gas law
here p is pressure and v is volume and t is temperature
so put here value and find v1
V1 = 5.71 cm³
volume of the bubble just before it reaches the surface is 5.71 cm³
The frequency of the 10th harmonic is 800 Hz
Explanation:
The frequency of the nth-harmonic for the standing waves in a string is given by the equation
where
is the fundamental frequency of the string
In this problem, we are given the frequency of the 3rd harmonic:
Which can be rewritten in terms of the fundamental frequency
So we find :
Now that we have the fundamental frequency, we can find the frequency of the 10th harmonic, with n = 10 :
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