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puteri [66]
2 years ago
6

Particles q1 = +8.0 °C, 92 = +3.5 uc, and

Physics
1 answer:
WINSTONCH [101]2 years ago
8 0

Answer:

Explanation:

q₁ = 8 x 10⁻⁶ C , q₂ = 3.5 x 10⁻⁶ C , q₃ = -2.5  x 10⁻⁶ C

Force between charge = 9 x 10⁹ x q₁ q₂ / d²

Force on q₁ due to q₂

= 9 x 10⁹ x 8 x 10⁻⁶ x 3.5 x 10⁻⁶  /( .10 )²

= 25.2 N . This force will be negative as it will point towards left ( q₂ will repel q₁ )

Force on q₁ due to q₃

= 9 x 10⁹ x 8 x 10⁻⁶ x 2.5 x 10⁻⁶  /( .15 )²

= 8 N . This force will be negative as it will point towards left ( q₃ will attract  q₁ )

Net force = 25.2 N - 8 N

= 17.2 N .

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6 0
3 years ago
The strength of the force of friction between two solids is affected by:
natali 33 [55]

Answer:

option a

Explanation:

<em>Th</em><em>e</em><em> </em><em>phase </em><em>of </em><em>matter </em>

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2 years ago
Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E= 3.
Nana76 [90]

Answer:

\sigma_i=1.06*10^{-6}C

Explanation:

When the space is filled with dielectric, an induced opposite sign charge appears on each surface of the dielectric. This induced charge has a charge density related to the charge density on the electrodes as follows:

\sigma_i=\sigma(1-\frac{E}{E_0})

Where E is the eletric field with dielectric and E_0 is the electric filed without it. Recall that \sigma is given by:

\sigma=\epsilon_0E_0

Replacing this and solving:

\sigma_i=\epsilon_0E_0(1-\frac{E}{E_0})\\\sigma_i=8.85*10^{-12}\frac{C^2}{N\cdot m^2}*3.40*10^5\frac{V}{m}*(1-\frac{2.20*10^5\frac{V}{m}}{3.40*10^5\frac{V}{m}})\\\sigma_i=1.06*10^{-6}C

3 0
3 years ago
The total thermal energy needed to raise the temperature of this ice cube to 0.0 °C and completely melt the ice cube is 5848 J
USPshnik [31]

Answer:

0.018kg

Explanation:

Q=mlf

m=?,Lf=334000J/kg,Q=5848J

5848=mx334000

5848=334000m

divide both sides by 334000

334000m/334000=5848/334000

m=0.018kg

4 0
3 years ago
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