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skad [1K]
3 years ago
15

A student obtained a clean dry glass-stoppered flask. she weighed the flask and stopper on an analytical balance and found the t

otal mass to be 31.601 g. she then filled the flask with water and obtained a mass for the full stoppered flask of 60.735 g. from these data, and the fact that at the temperature of the used water its density is 0.9973 g/cm3 , find the volume of the stoppered flask
Physics
1 answer:
svetoff [14.1K]3 years ago
6 0
29.213 cm3


First, calculate the mass of the water used. You do this by subtracting the original mass of the flask from the combined mass of the water and flask, giving:60.735 g - 31.601 g = 29.134 g
So we now know we have 29.134 g of water. To calculate the volume of the flask, simply divide by the density of the water, giving:29.134 g / (0.9973 g/cm3) = 29.213 cm3
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A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con
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Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \  \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm

The magnitude of the electric field is now given as;

E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

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