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Flura [38]
2 years ago
8

Four point charges, each of magnitude 2.38 µC, are placed at the corners of a square 75.2 cm on a side. If three of the charges

are positive and one is negative, find the magnitude of the force experienced by the negative charge. The value of Coulomb’s constant is 8.98755 × 10^9 N · m^2/C^2 . Answer in units of N.
Physics
1 answer:
poizon [28]2 years ago
7 0

Answer:

The Electric Force on Negative Charge is 2.968 N

Explanation:

charge on each corner, q = 2.38 micro coulomb

Side of square, a = 75.2 cm

Coulombic constant, K = 8.98755 x 10^9 Nm²/C²

sides of the square are A,B,C and D

and all sides of a square are equal so

AB = BC = CD = DA = 75.2 cm = 0.752 m

Diagonal, AC = BD = 1.414 x 0.752 = 1.06 m

Electric field at D due to charge at A

EA= Kq÷AB^2

= 8.98755×10^9 × 9.87×10^-6 ÷ 0.752^2

EA= 156863.82 N/C

Similarly Electric field at D due to charge C

EC=Kq÷CD^2

= 8.98755×10^9 ×9.87×10^-6 ÷ 0.752^2

EC= 156863.82 N/C

Electric field at D due to charge at BB

EB=Kq÷BD^2

EB=8.98755×10^9 × 9.87×10^-6 ÷ 1.06^2

EB=78949.01 N/C

Resolve the compoents

Ex = EA + EB cos 45

Ex = 156863.82 + 78949.01 x 0.707

Ex = 212689.2 N/C

Ey = EC + EB Sin 45

Ey = 156863.82 + 78949.01 x 0.707

Ey = 212689.2 N/C

The resultant electric field is

E = 1.414 x 212689.2 = 300787.95 N/C

the electric force on the negative charge is

F = q x E

F = 9.87 x 10^-6 x 300787.95

F = 2.968 N

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Answer:

The change in energy of the gas during the process is -1.83\times 10^{2} joules.

Explanation:

We can represent this process by the First Law of Thermodynamics, in which gas does work on its surroundings and absorbs heat from there to describe its change in energy. In other words:

Q_{in} - W_{out} = \Delta E

Where:

Q_{in} - Heat absorbed by the gas, measured in joules.

W_{out} - Work done by the gas, measured in joules.

\Delta E - Change in energy, measured in joules.

If we know that Q_{in} = 1.36\times 10^{2}\,J and W_{out} = 3.19\times 10^{2}\,J, the change in energy of the gas is:

\Delta E = 1.36\times 10^{2}\,J-3.19\times 10^{2}\,J

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3 years ago
When the distance between two masses is doubled, the gravitational attraction between
mrs_skeptik [129]

The new gravitational attraction will be 1/4 as much

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5 0
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17. A volleyball weighs about 300 grams.
atroni [7]

Answer:

PE = 44.1 J

Explanation:

Ok, to have the specific data, the first thing we must do is convert from grams to kilograms. Since mass must always be in kilograms (kg)

We have:

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We convert it using a rule of 3, replacing, simplifying units and solving:

  • \boxed{\bold{x=\frac{gr*1\ kg}{1000\ gr}=\frac{300\ gr*1\ kg}{1000\ gr}=\frac{300\ kg}{1000}=\boxed{\bold{0.3\ kg}}}}

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Earth's gravity is known to be 9.8 m/s², so we have:

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Use formula of potencial energy:

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Replace and solve:

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Since the decimal number, that is, the number after the comma is less than 5, it cannot be rounded, then we have this result.

The potential energy of the volleyball is <u>44.1 Joules.</u>

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