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Alja [10]
2 years ago
12

A sled slides down a hill with friction between the sled and hill but negligible alr resistance. Which of the following must be

correct about the resulting change in energy of the sled Earth system?
A) The sum of the kinetic energy and the gravitational potential energy changes by an amount equal to the energy dissipated by friction,
B)The gravitational potential energy decreases and the kinetic energy is constant.
C)The decrease in the gravitational potential energy is equal to the increase in kinetic energy
D)The gravitational potential energy and the kinetic energy must both decrease.
Physics
1 answer:
Andreyy892 years ago
8 0

Answer:

A) The sum of the kinetic energy and the gravitational potential energy changes by an amount equal to the energy dissipated by friction,

Explanation:

  • The kinetic energy is the energy that the object has and is defied by the work that is needed to accelerate the body.
  • The gravitational potential is a mechanism by which an equal amount of energy is being transferred per unit mass that is needed for the object to move from the specific location.
  • Hence when the sled moves down the hill with the force of gravity it has negligible resistance as an equal amount of energy is dissipated.
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Answer:

Momentum, p=\hbar k

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Wave A has a high frequency and a short wavelength. Wave B has a lower frequency and a longer wavelength. Which wave carries gre
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3 years ago
Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32.1 kj/mol.
Novosadov [1.4K]

Answer:

389.78681 K

Explanation:

P_1 = Initial pressure = 55.1 mmHg

P_2 = Final pressure = 1 atm = 760 mmHg

T_2 = Boiling point

T_1 = Initial temperature = 35°C

\Delta H_{vap} = Heat of vaporization = 32.1 kJ/mol

From the Clausius-Claperyon equation

ln\dfrac{P_2}{P_1}=(-\dfrac{\Delta H_{vap}}{R})(\dfrac{1}{T_2}-\dfrac{1}{T_1})\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{P_2}{P_1}\dfrac{R}{\Delta H_{vap}}+\dfrac{1}{T_1}\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{760}{55.1}\dfrac{8.314}{32.1\times 10^{3}}+\dfrac{1}{273.15+35}\\\Rightarrow T_2=\left(-ln\left(\frac{760}{55.1}\right)\frac{8.314}{32.1\times \:10^3}+\frac{1}{273.15+35}\right)^{-1}\\\Rightarrow T_2=389.78681\ K

The normal boiling point of the substance is 389.78681 K

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3 years ago
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