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Helga [31]
3 years ago
10

Capacitor 2 has half the capacitance and twice the potential difference as capacitor 1. What is the ratio (U_{\rm C})_1/\,(U_{\r

m C})_2.
Physics
1 answer:
IrinaK [193]3 years ago
4 0

Answer:

1/2

Explanation:

The energy stored in a capacitor is given by

U=\frac{1}{2}CV^2

where

C is the capacitance

V is the potential difference

Calling C_1 the capacitance of capacitor 1 and V_1 its potential difference, the energy stored in capacitor 1 is

U=\frac{1}{2}C_1 V_1^2

For capacitor 2, we have:

- The capacitance is half that of capacitor 1: C_2 = \frac{C_1}{2}

- The voltage is twice the voltage of capacitor 1: V_2 = 2 V_1

so the energy stored in capacitor 2 is

U_2 = \frac{1}{2}C_2 V_2^2 = \frac{1}{2}\frac{C_1}{2}(2V_1)^2 = C_1 V_1^2

So the ratio between the two energies is

\frac{U_1}{U_2}=\frac{\frac{1}{2}C_1 V_1^2}{C_1 V_1^2}=\frac{1}{2}

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6 0
3 years ago
Which substance(s) will sink if placed in a fluid that has a density of 1.19 g/ml? (List all)
olga2289 [7]

Answer:

Corn Syrup

Explanation:

7 0
3 years ago
Two engineering students, John with a weight of 92 kg and Mary with a weight of 46 kg, are 30 m apart. Suppose each has a 0.04%
Hoochie [10]

Answer:

F = 6.27 x 10 ¹⁹ N

Explanation:

Given

m₁ = 92 kg, m₂ = 46 kg, % = 0.04% N = 6.022 x 10²³ Z = 18, e = 1.6 x 10 ⁻¹⁹ C, M = 0.018 kg/mol

q₁ = % * [m * N * A * e / M ]  

q₁ = 0.0004 * [ ( 92 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]

q₁ = 3.54 x 10⁶ C

q₂ = 0.0004 * [ ( 46 kg * 6.022 x 10²³ * 18 * 1.6 x 10 ⁻¹⁹ ) / (0.018 kg/mol ) ]

q₂ = 1.773 x 10⁶ C

Now to determine the electrostatic force con use the equation

F = K * q₁ * q₂ / d²

K = 8.99 x 10 ⁹

F = 8.99 x 10 ⁹ * 3.54 x 10⁶ C * 1.773 x 10⁶ C / (30m)²

F = 6.27 x 10 ¹⁹ N

3 0
3 years ago
A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of
sladkih [1.3K]

Answer:

The magnitude of the resultant of the magnetic field is 4.11\times10^{-5}\ T

Explanation:

Given that,

Current = 40 A

Magnetic field B=3.7\times10^{-5}\ T

Distance = 22 cm

We need to calculate the magnetic field

Using formula of magnetic field

B'=\dfrac{\mu_{0}I}{2\pi r}

Where, r = distance

I = current

Put the value into the formula

B'=\dfrac{4\pi\times10^{-7}\times20}{2\pi\times0.22}

B'=1.8\times10^{-5}\ T

We need to calculate the magnitude of the resultant of the magnetic field

Using formula of resultant

B''=\sqrt{B^2+B'^2}

Put the value into the formula

B''=\sqrt{(3.7\times10^{-5})^2+(1.8\times10^{-5})^2}

B''=4.11\times10^{-5}\ T

Hence, The magnitude of the resultant of the magnetic field is 4.11\times10^{-5}\ T

6 0
3 years ago
In a particular run of the “Archimedes principle” experiment a particular unknown substance is found to have a “dry mass” of 4.5
iris [78.8K]

Answer:

D = 18000 kg/m3

V = 2.5*10{-7}m3

Explanation:

From the Archimedes principle,

Weight of fluid displaced = W_{air} - W_{water}

W_{air} = 4.5 gm

W_{water} = 4.25 gm

W = [4.5 - 4.25]*9.81*10^{-3}

W = 2.4525*10{-3} N

\frac{density\ of\ object}{density\ of\ fluid} = \frac{weight\ in\ air}{weight\ of\ displaced\ fluid}

Density\ of\ object = \frac{D_{water}*Weight\ in\ air}{weight\ of\ displaced\ water}

D = \frac{1000*4.5*10^{-3}*9.8}{2.4525*10^{-3}}N

D = 18000 kg/m3

b) object Volume can be obtained  as ,

V = \frac{m}{D} = \frac{4.5*10^{-3}}{18000}

V = 2.5*10{-7}m3

7 0
3 years ago
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