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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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A compound whose molecules contain one boron atom and three fluorine atoms would be named monoboron fluoride. Please select the
Slav-nsk [51]
I know it is false...............

5 0
3 years ago
31. Draw a free body diagram for a 15.5N box that is being pushed to the right with a 18. N force while experiencing 4.30 N of r
posledela

Answer:

See answers below

Explanation:

a.

F = mg,

15.5 N = m(9.8 m/s²)

m = 1.58 kg

b.

Fnet = Applied force - resistance,

Fnet = 18 N - 4.30 N,

Fnet = 13.70 N

Fnet = ma

13.70 N = (1.58 kg)a

a = 8.67 m/s²

For the free body diagram, draw a box with an upward arrow labeled 15.5 N, a downward label labeled 15.5 N, a right label labeled 18 N, and a left label labeled 4.30 N.

7 0
2 years ago
A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at
Vesna [10]

Answer:

Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t

Explanation:

Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K

mg=k x

1 x 32= k x 2

K=16

And also give that damping force is 8 times the velocity so damping constant C=8.

We know that equation for spring mass system

my''+Cy'+Ky=F

Now by putting the values

1 y"+8 y'+ 16y=6 cos 4 t ----(1)

The general solution of equation Y=CF+IP

Lets assume that at steady state the equation of y will be

y(IP)=A cos 4t+ B sin 4t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -4A sin 4t+4B cos 4t

y"=-16A cos 4t-16B sin 4t

Now put the values of y" , y' and y in equation 1

1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t

So by comparing the coefficient both sides

-16A+32B+16A=0  So B=0

-16 B-32 A+16B=6  So A=-3/16

y=-3/16 cos 4t

Now to find the CF  of differential equation 1

y"+8 y'+ 16y=6 cos 4 t

Homogeneous version of above equation

m^2+8m+16=0

So CF =(C_1+tC_2)e^{-2t}

So the general equation

Y=(C_1+tC_2)e^{-2t}-3/16 cos 4t

Given that t=0 Y=0 So

C_1=\dfrac{3}{16}

t=0 Y'=0 So

C_2 =\dfrac{3}{8}

Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t

The above equation is the general equation for motion.

3 0
3 years ago
What is the magnitude of the sum of the two vectors A = 36 units at 53 degrees, and B =47 units at 157 degrees.
Thepotemich [5.8K]

Answer:

51.82

Explanation:

First of all, let's convert both vectors to cartesian coordinates:

Va = 36 < 53° = (36*cos(53), 36*sin(53))

Va = (21.67, 28.75)

Vb = 47 < 157° = (47*cos(157), 47*sin(157))

Vb = (-43.26, 18.36)

The sum of both vectors will be:

Va+Vb = (-21.59, 47.11)   Now we will calculate the module of this vector:

|Va+Vb| = \sqrt{(-21.59)^2+(47.11)^2}=51.82

4 0
4 years ago
A diffraction grating contains 15,000 lines/inch. We pass a laser beam through the grating. The wavelength of the laser is 633 n
MatroZZZ [7]

Answer:

Recall the Diffraction grating formula for constructive interference of a light

y = nDλ/w                                      Eqn 1

Where;

w = width of slit = 1/15000in =6.67x10⁻⁵in =   6.67x10⁻⁵ x 0.0254m = 1.69x10⁻⁶m

D = distance to screen  

λ = wavelength of light  

n = order number  = 1

Given  

y1 = ? from 1st order max to the central  

D = 2.66 m  

λ = 633 x 10-9 m  

and n = 1  

y₁ = 0.994m  

Distance (m) from the central maximum (n = 0) is the first-order maximum (n = 1)                =         0.994m

Q b. How far (m) from the central maximum (m = 0) is the second-order maximum (m = 2) observed?

w = width of slit = 1/15000in =6.67x10⁻⁵in =   6.67x10⁻⁵ x 0.0254m = 1.69x10⁻⁶m

D = distance to screen  

λ = wavelength of light  

n = order number  = 1

Given  

y1 = ? from 1st order max to the central  

D = 2.66 m  

λ = 633 x 10⁻⁹ m  

and n =  2

y₂ = 0.994m  

Distance (m) from the central maximum (n = 0) is the first-order maximum (n = 2) =1.99m

8 0
3 years ago
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