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Degger [83]
4 years ago
10

The charge on a parallel plate capacitor is 7.68 C, and the plate is connected to a 9.0 V battery. What is the capacitance of th

e capacitor?
A. 8.5x10^-7 pF
B. 1.1x10^-6 pF
C. 8.5x10^5 pF
D. 6.9x10^7 pF
Physics
1 answer:
Natali5045456 [20]4 years ago
8 0

Charge on a capacitor = (capacitance) x (voltage)

7.68 Coulomb = (capacitance) x (9.0 V)

Capacitance = (7.68 Coul) / (9.0 V)

<em>Capacitance = 0.83 Farad</em>

<em></em>

Apparently, the " 10^something " after the 7.68 C got lost.

If the charge is actually 7.68 <u>micro</u>coulombs, then the answer is<em> (A) .</em>

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The chart below shows the dates of eclipses in the years 2006 and 2007. What is the best explanation for the frequency of eclips
xz_007 [3.2K]

Answer:

It's A

As the Gizmo shows, the Moon orbits at an angle relative to the plane of Earth's orbit. Usually, the Moon is above or below the Earth when an eclipse could happen. Every six months or so, the Moon, Earth and Sun are lined up sufficiently to create a lunar and a solar eclipse.

4 0
3 years ago
A population of mammals leaves an ecosystem because of increased temperatures and decreased rainfall. What's the reason for this
DochEvi [55]

Answer:

A. change in abiotic factors

Explanation:

Abiotic factors are those factors that have no life, the most important of which can be mentioned: water, temperature, light, pH, soil, humidity, oxygen and nutrients.

3 0
4 years ago
In a charging process, 1 × 10^13 electrons are removed from one small metal sphere and placed on a second identical sphere. Init
liraira [26]

Answer:

r = 0.303m

= 30.3cm

Explanation:

Given that,

The number of electrons transferred from one sphere to the other,  

n  = 1 ×10 ¹³e le c t r o n s

The electrostatic potential energy between the spheres,  

U = − 0.061 J

The charge on an electron,  

q = − 1.6 × 10 ⁻¹⁹C

The coulomb constant,  

K = 8.98755 × 10 ⁹ N ⋅ m ² / C 2²

Due to the transfer of electrons, both spheres become equally and oppositely.

The charge gained by the sphere due to the excess of the electron is:  

q ₁ = n q

   = 1 ×10 ¹³ *  − 1.6 × 10 ⁻¹⁹

   = -1.6  × 10⁻⁶C

The charge left on the first sphere is =

q ₂ = -q₁ = 1.6  × 10⁻⁶C

The electric potential energy between two point charges is given by the following equation:

U = K q ₁q ₂/r

q ₁ and  q ₂ are the two charges.

r  is the distance between the charge and the point.

K  =  8.98755  ×  10 ⁹ N ⋅ m ² / C ²

we have:

-0.061 =  (8.98755  ×  10 ⁹ * (-1.6  × 10⁻⁶)²) / r

r = (18.41 ×  10 ⁻³) / 0.061

r = 0.303m

= 30.3cm

4 0
4 years ago
A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi
Nataliya [291]

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

m_{m} = mass of metal = 400 g

c_{m} = specific heat of metal = ?

T_{mi} = initial temperature of metal = 100 °C

m_{a} = mass of aluminum cup = 100 g

c_{a} = specific heat of aluminum cup = 900.0 J/kg ∙ K

T_{ai} = initial temperature of aluminum cup = 15 °C

m_{w} = mass of water = 500 g

c_{w} = specific heat of water = 4186 J/kg ∙ K

T_{wi} = initial temperature of water = 15 °C

T = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

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Answer:

pelvic gridle

Explanation:

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