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4 years ago

Would a decrease in force cause a decrease in pressure

Physics

1 answer:

natta225 [31]4 years ago

4 0

IdkAnswer:

Explanation:

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To see how two traveling waves of the same frequency create a standing wave.

WARRIOR [948]

Answer: The wave is traveling in the - x direction.

Explanation: The parameter in a wave function determines the direction of the wave is "ωt"

Where ω = angular frequency(in hertz ) and t = time taken (in seconds)

The product of ωt = 2π which is angular displacement in radian.

A negative value of the of ωt means the wave is traveling in the negative direction.

Also a positive value of sin ωt means the wave is traveling in the positive direction

7 0

3 years ago

Water flows past a flat plate that is oriented parallel to the flow with an upstream velocity of 0.4 m/s. (a) Determine the appr

Trava [24]

Answer:

1.12 m

0.08291 m

Explanation:

u = Upstream velocity = 0.4 m/s

Re = Reynold's number = $5\times 10^5$ (turbulent)

$\nu$ = Viscosity of water = $1.12\times 10^{-6}\ Pas$

Here the flow is turbulent so we have the relation

$Re_{xcr}=\frac{ux_{cr}}{\nu}\\\Rightarrow x_{cr}=\frac{Re_{xcr}\nu}{u}\\\Rightarrow x_{cr}=\frac{5\times 10^5\times 1.12\times 10^{-6}}{0.4}\\\Rightarrow x_{cr}=1.4\ m$

The approximate location downstream from the leading edge where the boundary layer becomes turbulent is 1.4 m

Boundary layer thickness relation is given by

$\delta={\frac{\nu x}{u}}^{\frac{1}{5}}\\\Rightarrow \delta={\frac{1.12\times 10^{-6}\times 1.4}{0.4}}^{\frac{1}{5}}\\\Rightarrow \delta=0.08291\ m$

The boundary layer thickness is 0.08291 m

4 0

4 years ago

It takes 105 j of work to move 2.7 c of charge from the negative plate to the positive plate of a parallel plate capacitor. what

wolverine [178]

Se necesita 105 j de trabajo para mover 2,7 c de carga de la placa negativa a la placa positiva de un condensador de placa paralela. Qué diferencia de tensión existe entre las placas

5 0

3 years ago

A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy

SashulF [63]

Answer:

$5.444\times 10^{-4}$

Explanation:

The momentum of the neutron before and after the decay is the same since there's no external force.

$P_{sys}=const\\\\P=mv\\\\K=0.5mv^2$

#The neutron is initially at rest, so after the decay:

$P_A+P_B=0\\\\P_A=-P_B$

#After decay, the proton has +ve direction with a velocity $v_A$ while the electron moves in a negative direction with a velocity $v_B$

Therefore:

$P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B$

Let the energy released during the decay be Q:

$Q=K_{tot}=K_A+K_B\\\\Q=K_A+0.5m_Bv_B^2\\\\Q=K_A+0.5m_B(\frac{m_A}{m_B})^2v_A^2\\\\\ But \ K_A=0.5m_Av_A^2\\\\\therefore Q=K_A+\frac{m_A}{m_B}K_A=K_A(1+\frac{m_A}{m_B})\\\\=Q=\frac{m_A+m_B}{m_B}K_A\\\\m_A=1836m_B\\\\\frac{K_A}{Q}=\frac{m_B}{1836m_B+m_B}=\frac{1}{1837}\\\\\frac{K_A}{Q}=5.444\times10^{-4}$

Hence,Kp/Ktot is 5.444x10^(-4)

4 1

3 years ago

Discuss Joule-Thompson effect with relevant examples and formulae.

Delicious77 [7]

Answer:

$\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp$

Explanation:

Joule -Thompson effect

Throttling phenomenon is called Joule -Thompson effect.We know that throttling is a process in which pressure energy will convert in to thermal energy.

Generally in throttling exit pressure is low as compare to inlet pressure but exit temperature maybe more or less or maybe remains constant depending upon flow or fluid flow through passes.

Now lets take Steady flow process

Let

$P_1,T_1$ Pressure and temperature at inlet and