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iragen [17]
4 years ago
8

A 75-kilogram bicyclist coasts down a hill at a constant speed of 12 meters per second. What is the kinetic

Physics
2 answers:
MA_775_DIABLO [31]4 years ago
8 0

Answer:

(3) 5.4 × 10^3 J

Explanation:

The kinetic energy of the bicyclist can be calculated using the following equation:

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

where

m = 75 kg is the mass of the bicyclist

v = 12 m/s is the speed of the byciclist

By substituting these numbers into the equation, we find:

K=\frac{1}{2}mv^2 = \frac{1}{2}(75 kg)(12 m/s)^2=5400 J=5.4 \cdot 10^3 J

So, the correct answer is

(3) 5.4 × 10^3 J

Anestetic [448]4 years ago
6 0
Kinetic E=1/2mv^2

KE = 1/2 * 75 * 12^2 = 5400J = 5.4*10^3 J


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Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat
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Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

                               j_{H_2O}=\frac{D}{l} \bigtriangleup c

where

  • j_{H_2O} is the mass flux of the water which is to be calculated.
  • D is diffusion coefficient which is given as 0.25 cm^2/s
  • l is the thickness of the film which is 0.15 cm thick.
  • \bigtriangleup c is given as

                                \bigtriangleup c= \frac{P_{sat}-P_a}{RT}

In this

  • P_{sat} is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa
  • P_a is the air pressure which is given as 0.5 times of P_{sat}
  • R is the universal gas constant as 8.314 kJ/kmol-K
  • T is the temperature in Kelvin scale which is 20+273= 293K

By substituting values in the equation

                                    \bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3

Converting \bigtriangleup c into cm^3/cm^3

As 1 mole of water 18 cm^3 so

                               \bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6}  cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6}  cm^3/cm^3

Putting this in the equation of mass flux equation gives

                            j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6}  cm/s

For calculation of water level drop in a day, converting mass flux as

                     j_{H_2O}=14.4 \times 10^{-6}  \times 24 \times 3600  cm/day\\ j_{H_2O}=1.24  cm/day

So the water level will drop by about 1.24 cm in 1 day.

7 0
4 years ago
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