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4 years ago

Is Newton's third law of motion relevant to dropping an egg

1 answer:

8 0

Newton's third law states, that for every action, there is an equal and opposite reaction. In this case, the third law of motion is relevant to the egg being dropped, as the forces for gravity are pulling downwards, the air drag is pushing up on the egg. As well as the force of the ground when the egg lands.

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Suppose a plot of inverse wavelength vs frequency has slope equal to 0.119, what is the speed of sound traveling in the tube to

strojnjashka [21]

Answer:

8.40 m/s

Explanation:

Slope of the plot is 0.119

Slope of a plot is given by the change in y direction divided by the change in x direction

Here, the y axis represents inverse wavelength and the x axis represents frequency.

f = Frequency (Hz, assumed)

v = Phase velocity (m/s, assumed)

λ = Wavelength (m, assumed)

So, slope

$m=\frac{\frac{1}{\lambda}}{f}$

Now,

$\lambda=\frac{v}{f}\\\Rightarrow \lambda^{-1}=\frac{f}{v}$

$\\\Rightarrow m=\frac{\frac{f}{v}}{f}\\\Rightarrow 0.119=\frac{1}{v}\\\Rightarrow v=\frac{1}{0.119}\\\Rightarrow v=8.40\ m/s$

The speed of sound travelling in the tube is 8.40 m/s

5 0

4 years ago

Which wavelength would scientists use to measure the molecular structure of H2O?

Ne4ueva [31]

Answer:

To find out what water is made of, it helps to look at its chemical formula, which is H2O. This basically tells us that the water molecule is composed of two elements: hydrogen and oxygen or, more precisely, two hydrogen atoms (H2) and one oxygen atom (O).

Explanation:

7 0

3 years ago

Which dog has the most kinetic energy

Blababa [14]

The formula to find the kinetic energy is:

Ek= 1/2 × m × v^2

1. Ek= 1/2×15×3^2
= 67.5 J

2.Ek= 1/2×8×4^2
=64 J

3.Ek= 1/2×12×5^2
= 150 J

4.Ek= 1/2×10×6^2
= 180 J

So the fourth dog has the most kinetic energy.

4 0

3 years ago

Determine the power that needs to besupplied by the fanifthe desired velocity is 0.05 m3/s and the cross-sectional area is 20 cm

Mariulka [41]

Answer:

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

Explanation:

Complete statement is: <em>Determine the power that needs to besupplied by the fan if the desired velocity is 0.05 cubic meters per second and the cross-sectional area is 20 square centimeters.</em>

From Thermodynamics and Fluid Mechanics we know that fans are devices that work at steady state which accelerate gases (i.e. air) with no changes in pressure. In this case, mechanical rotation energy is transformed into kinetic energy. If we include losses due to mechanical friction, the Principle of Energy Conservation presents the following equation:

$\eta\cdot \dot W = \dot K$

$\dot W = \frac{\dot K}{\eta}$ (Eq. 1)

Where:

$\eta$ - Efficiency of fan, dimensionless.

$\dot W$ - Electric power supplied fan, measured in watts.

$\dot K$ - Rate of change of kinetic energy of air in time, measured in watts.

From definition of kinetic energy, the equation above is now expanded:

$\dot W = \frac{\rho_{a}\cdot \dot V}{2\cdot \eta}\cdot \left(\frac{\dot V}{A_{s}} \right)^{2}$ (Eq. 2)

Where:

$\rho_{a}$ - Density of air, measured in kilograms per cubic meter.

$\dot V$ - Volume flow, measured in cubic meters per second.

$A_{s}$ - Cross-sectional area of fan, measured in square meters.

If we know that $\rho_{a} = 1.20\,\frac{kg}{m^{3}}$ , $\dot V = 0.05\,\frac{m^{3}}{s}$ , $\eta = 0.3$ and $A_{s} = 20\times 10^{-4}\,m^{2}$ , the power needed to be supplied by the fan is:

$\dot K = \left[\frac{\left(1.20\,\frac{kg}{m^{3}} \right)\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{2\cdot (0.3)} \right]\cdot \left(\frac{0.05\,\frac{m^{3}}{s} }{20\times 10^{-4}\,m^{2}} \right)^{2}$

$\dot K = 62.5\,W$

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

5 0

3 years ago

Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te

seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy 5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant $\gamma =1.4$