When the temperature of a gas in a balloon is reduced by half, the average kinetic energy of the particles is

Physics

1 answer:

Aneli [31]3 years ago

8 0

According to KE = (3/2)kT

reducing temperature, in KELVIN, by half, average KE is reduced by half.

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How much heat is needed to melt 13.74 kg of silver that is initially at 20°C?

salantis [7]

Answer:

$Q = 4.63 \times 10^6 J$

Explanation:

As we know that melting point of silver is

T = 961.8 degree C

Latent heat of fusion of silver is given as

L = 111 kJ/kg

specific heat capacity of silver is given as

$s = 240 J/kg C$

now we will have

$Q = ms\Delta T + mL$

$\Delta T = 961.8 - 20$

$\Delta T = 941.8 degree C$

now from above equation

$Q = (13.74)(240)(941.8) + (13.74)(111 \times 10^3)$

$Q = 3.1 \times 10^6 + 1.53 \times 10^6$

$Q = 4.63 \times 10^6 J$

3 0

3 years ago

A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci

MA_775_DIABLO [31]

Answer:

$v_B=3.78\times 10^5\ m/s$

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is $6.63\times 10^{-27}\ kg$

Speed of nucleus at A is $v_A=6.2\times 10^5\ m/s$

Potential at point A, $V_A=1.5\times 10^3\ V$

Potential at point B, $V_B=4\times 10^3\ V$

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

$\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s$