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Cloud [144]
3 years ago
5

Write an equation that expresses the following relationship. u varies jointly with p and d and inversely with w In your equation

, use k as the constant of proportionality.
Physics
1 answer:
Alekssandra [29.7K]3 years ago
5 0
<h2>Answer:</h2>

\boxed{y=k\frac{pd}{w}}

<h2>Explanation:</h2>

Let's explain what direct and indirect variation mean:

  • When we say that y varies jointly as x \ and \ w, we mean that:

y=kxw for some nonzero constant k that is the constant of variation or the constant of proportionality.

  • On the other hand, when we say that y varies inversely as x or y is inversely proportional to x, we mean that:

y=\frac{k}{x} for some nonzero constant k, where k is also the constant of variation.

___________________

In this problem, u varies jointly with p and d and inversely with w, being k the constant of proportionality, then:

\boxed{y=k\frac{pd}{w}}

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Find the intensity of a 55 dB sound given I 0=10^-12W/m^2
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Answer:

3.16 × 10^{-7} W/m^{2}

Explanation:

β(dB)=10 × log_{10}(\frac{I}{I_{0} })

I_{0}=10^{-12} W/m^{2}

β=55 dB

Therefore plugging into the equation the values,

55=10 log_{10}(\frac{I}{ [tex]10^{-12}})[/tex]

5.5= log_{10}(\frac{I}{ [tex]10^{-12}})[/tex]

10^{5.5}= \frac{I}{10^{-12} }

316227.76×10^{-12}= I

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3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

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Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

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Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

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Answer:

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Explanation:

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