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3 years ago

6

Consider the following mass distribution where the x- and y-coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 2.5 kg at (

0.0, 3.3) m, and 4.0 kg at (2.8, 0.0) m. Where should a fourth object of 7.1 kg be placed so that the center of gravity of the four-object arrangement will be at (0.0, 0.0) m?

1 answer:

nekit [7.7K]3 years ago

8 0

Answer: x = -1.6 y= -1.2

Explanation:

By definition the x- and –y coordinates of the center of mass for a discrete set of bodies, are given by the following expressions:

Xi = ∑ mi. xi / ∑ mi

Yi = ∑ mi. yi / ∑ mi

In order to the x- and –y coordinates of the center of mass of all the system, be (0, 0),

both num.erators in the Xi, Yi expressions must be equal to 0.

So, replacing by the values, and taking into account that the 5.0 Kg is located in the origin, we can write the following:

Xi = 2.8 . 4 Kg + X4 . 7.1 Kg = 0 → X4 = -1.6

Yi = 3.3 . 2.5 Kg + Y4 . 7.1 Kg = 0 → Y4 = -1.2

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PV=nRT

Here

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$\\ \tt\bull\leadsto nRT$

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3 years ago

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irakobra [83]

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2 years ago

Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y

WITCHER [35]

Question:

A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity $3.65\times 10^5N/C$

Explanation:

A point charge ,q = $-2.14\times 10^{-6} C$ is located in the center of a spherical cavity of radius , $r =6.55\times 10^{-2}$ m inside an insulating spherical charged solid.

The charge density in the solid , d = $7.35 \times 10^{-4}C/m^3.$

Distance from the center of the cavity,R = $9.5\times 10^{-2 }m$

Volume of shell of charge= V = $(\frac{4\pi}{3})[ R^3 - r^3 ]$

Charge on the shell ,Q = $V \times d'$

$Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d$

$Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}$

$Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}$

$Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}$

$Q =1.7745 \times 10^{-6 }C$

Electric field at $9.5\times 10^{-2}$ m due to shell $E1 = \frac{k Q}{R^2}$

E1 = $\frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E1 =1.769\times 10^6 N/C$

Electric field at $9.5\times 10^{-2}$ due to 'q' at center $E2 = \frac{kq}{R^2}$

E2 = $\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E2 =2.134\times 10^6 N/C$

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

$=[ 2.134 - 1.769 ]\times 10^6$

$= 3.65\times 10^5 N/C$

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