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Amiraneli [1.4K]

2 years ago

5

Charge q1 is distance r from a positive point charge q. charge q2=q1/3 is distance 2r from q. what is the ratio u1/u2 of their p

otential energies due to their interactions with q?

1 answer:

makvit [3.9K]2 years ago

5 0

We need the power law for the change in potential energy (due to the Coulomb force) in bringing a charge q from infinity to distance r from charge Q. We are only interested in the ratio U₁/U₂, so I'm not going to bother with constants (like the permittivity of space).

<span>The potential energy of charge q is proportional to </span>
<span>∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹, </span>

<span>so if r₂ = 3r₁ and q₂ = q₁/4, then </span>
<span>U₁/U₂ = q₁Qr₂/(r₁q₂Q) = (q₁/q₂)(r₂/r₁) </span>
<span>= 4•3 = 12.</span>

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snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

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And the mass flow rate is

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Mass flow rate in the outlet=density*area*outlet velocity*time

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(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

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As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

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$Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s$

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Then the inlet volume will be

$Inlet volume = inlet velocity*area of circle=\pi r^{2}*inlet velocity$

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

$Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}$

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2 years ago

A ball was kicked upward at a speed of 64.2 m/s. how fast was the ball going 1.5 seconds later

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second, on account of gravity.

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= (64.2 m/s) - [ 14.7 m/s ]

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2 years ago

An electric motor consumes 9.00 kj of electrical energy in 1.00 min. if one-third of this energy goes into heat and other forms

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2 years ago

A material that has a fracture toughness of 33 MPa.m0.5 is to be made into a large panel that is 2000 mm long by 250 mm wide and

scoray [572]

Answer:

$F_{allow} = 208.15kN$

Explanation:

The word 'nun' for thickness, I will interpret in international units, that is, mm.

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$K_{allow} =\frac{K_c}{N}$

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$K_{allow} = \frac{33Mpa*m^{0.5}}{2} = 16.5MPa.m^{0.5}$

Now we will need to find the average width of both the crack and the panel, these values are found by multiplying the measured values given by 1/2

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$\alpha = 0.5*L_c = 0.5*4mm = 2mm$

<em>For the panel</em>

$\gamma = 0.5*W = 0.5*250mm = 125mm$

To find now the goemetry factor we need to use this equation

$\beta = \sqrt{sec(\frac{\pi\alpha}{2\gamma})}\\\beta = \sqrt{sec(\frac{2\pi}{2*125mm})}\\\beta = 1$

That allow us to determine the allowable nominal stress,

$\sigma_{allow} = \frac{K_{allow}}{\beta \sqrt{\pi\alpha}}$

$\sigma_{allow} = \frac{16.5}{1*\sqrt{2*10^{-3} \pi}}$

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So to get the force we need only to apply the equation of Force, where

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2 years ago

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Yanka [14]

Answer:

Diameter of the cylinder will be $d=2.998\times 10^4m$

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Change in length $\Delta l=0.38mm$

Length of rod $l=500mm$

Load F = 11100 KN

Strain is given by $strain=\frac{\Delta l}{l}=\frac{0.38}{500}=7.6\times 10^{-4}$

We know that young's modulus $E=\frac{stress}{strain}$

So $207\times 10^9=\frac{stress}{7.6\times 10^{-4}}$

$stress=1573.2\times 10^{-5}N/m^2$

We know that stress $=\frac{force}{artea }$

So $1573.2\times 10^{-5}=\frac{11100\times 1000}{area}$

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6 0

2 years ago