Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
Elements is the answer to this question
The wavelength of the golf ball is <u>2.328×10⁻³⁴m.</u>
All moving particles with mass have a matter wave associated with it. These matter waves are called deBroglie waves.
The deBroglie wavelength λ of a particle is given by,

Here, h is the Planck's constant, m is the mass of the ball and v is its velocity.
Calculate the deBroglie wavelength of the moving golf ball by substituting 6.626×10⁻³⁴J s for h, 45.9×10⁻³kg for m and 62.0 m/s for v.

The wavelength of the golf ball is <u>2.328×10⁻³⁴m.</u>
8 electron are needed for bonding