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Sever21 [200]
4 years ago
12

An ideal gas initially at 275 k undergoes an isobaric expansion at 2.50 kpa. the volume increases from 1.00 m3 to 3.00 m3 and 11

.2 kj is transferred to the gas by heat. (a) what is the change in internal energy of the gas?
Physics
1 answer:
tekilochka [14]4 years ago
5 0
The transformation is isobaric (constant pressure), so the work done by the gas is the product between its pressure and the variation of volume:
W=p \Delta V=(2500 Pa)(3.0 m^3-1.0 m^2 ) = +5000  J =+5.0 kJ

The heat transferred to the gas is
Q=+11.2 kJ

So we can use the first law of thermodynamics to compute the variation of internal energy of the gas:
\Delta U = Q-W=11.2 kJ-5.0 kJ=+6.2 kJ = +6200 J
where the positive sign means the internal energy of the gas has increased.
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A mass spectrometer applies a voltage of 2.00 kV to accelerate a singly charged positive ion. A magnetic field of B = 0.400 T th
N76 [4]

The mass of the ion is 5.96 X 10⁻²⁵ kg

<u>Explanation:</u>

The electrical energy given to the ion Vq will be changed into kinetic energy \frac{1}{2}mv^2

As the ion moves with velocity v in a magnetic field B then the magnetic Lorentz force Bqv will be balanced by centrifugal force \frac{mv^2}{r}.

So,

Vq = \frac{1}{2}mv^2

and

Bqv = \frac{mv^2}{r}

Right from these eliminating v, we can derive

m = \frac{B^2r^2q}{2V}

On substituting the value, we get:

m = \frac{(0.4)^2X (0.305)^2 X1.602X 10^-^1^9}{2X 2000}\\\\

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3 0
3 years ago
A gold sphere of radius R=100 μm and density 19g/cm^3 falls through water. Given the viscosity of water is about 10^-3​ Pa s and
icang [17]

The terminal velocity of gold sphere is 39.2 cm/s

<h3>What is terminal velocity?</h3>

Terminal velocity is the maximum velocity attainable for an object as it falls through a fluid.

<h3>How to calculate the terminal velocity of the gold sphere?</h3>

The terminal velocity of the gold sphere is given by v = 2gr²(ρ - σ)/9η where

  • g = acceleration due to gravity = 9.8 m/s²,
  • r = radius of sphere = 100 μm = 100 × 10⁻⁶ m = 10⁻⁴ m = 10⁻² cm,
  • ρ = density of sphere = 19 g/cm³,
  • σ = density of water = 1.0 g/cm³ and
  • η = viscosity of water = 10⁻³ Pa-s

So, susbtituting the values of the variables into the equation, we have that

v = 2gr²(ρ - σ)/9η

v = 2 × 9.8m/s²× (10⁻² cm)²(19 g/cm³ - 1.0 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 9.8 m/s² × 10⁻⁴ cm² × (18 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 980 cm/s² × 10⁻⁴ cm² × 2 g/cm³/(1 × 10⁻³ Pa-s)

v = 3920 g/s² × 10⁻⁴/(1 × 10⁻³ Pa-s)

v = 392 cm/s × 10³ × 10⁻⁴

v = 392 × 10⁻¹ cm/s

v = 39.2 cm/s

So, the terminal velocity is 39.2 cm/s

Learn more about terminal velocity of sphere here:

brainly.com/question/21684177

#SPJ1

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