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madam [21]
3 years ago
8

A cylindrical capacitor is made of two thin-walled concentric cylinders. The inner cylinder has radius 8 mm , and the outer one

a radius 11 mm . The common length of the cylinders is 152 m . What is the potential energy stored in this capacitor when a potential difference 8 V is applied between the inner and outer cylinder? (k =1/4∗π∗ϵ0=8.99×109N⋅m2/C2) A cylindrical capacitor is made of two thin-walled concentric cylinders. The inner cylinder has radius 8 , and the outer one a radius 11 . The common length of the cylinders is 152 . What is the potential energy stored in this capacitor when a potential difference 8 is applied between the inner and outer cylinder? ( =) 8.5×10−7 J 7.5×10−6 J 2.0×10−7 J 1.4×10−7 J 9.6×10−7 J
Physics
1 answer:
yanalaym [24]3 years ago
3 0

Answer:

So the answer is the first one

E = 8.5 x 10 ⁻⁷ J

Explanation:

Given:

r₁ = 8 mm , r₂ = 11 mm , h = 152 m

ε₀ = 8.85 x 10⁻¹² N * m² / C²

C = Q / V

C = 2π * h * ε₀ / ln ( r₂ / r₁ )

C = [ 2π * 152 m * 8.85 x 10⁻¹² N * m² / C² ]  / [ ln ( 11 mm / 8mm ) ]

C = 2.654 x 10⁻⁸ F

C = 26.54 nF

Now to determine the energy Emf

E = ¹/₂ * C * V²

E = ¹/₂ * 26.54 nF * 8²V

E = 8.5 x 10 ⁻⁷ J

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max2010maxim [7]

Newton’s 2nd law states that Force is equal to the product of mass (m) and acceleration (a):

F = m a                                  ---> 1

While in magnetic forces, force can also be expressed as:

F = q v B                               ---> 2

where,

q = total charge

v = velocity = 45 cm / s = 0.45 m / s

B = the magnetic field = 85 T

First we solve for the total charge, q:

q = 3.8 × 10^-23 g (1 mol / 23 g) (6.022 × 10^23 electrons / mol) (1.602 × 10^-19 C / electron)

q = 1.594 × 10^-19 C

 

We equate equations 1 and 2 then solve for acceleration a:

m a = q v B

a = q v B / m

a = [1.594 × 10^-19 C * 0.45 m / s * 85 T] / 3.8 × 10-26 kg

a = 160,437,862.2 m/s^2

 

Therefore the maximum acceleration of Na ions is about 160 × 10^6 m/s^2.

5 0
3 years ago
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Suppose you want to make an interferometer with two slits to show the wave nature of light to your friends. The most convenient
nevsk [136]

Answer:

a) the pattern is bright,  b)  y = 8.25 mm, c) the distance increases with the wavelength increases

Explanation:

This is a team building exercise to perform interference experiments, where the separation of the slits is d = 1 m = 1 10⁻³ m and the distance to the screen L = 1.5 m  

Constructive interference occurs when the path difference is equal to an integer number of wavelengths  

             d sin θ = m λ  

a) The center of the two-ray pattern has the same length so the intensity is MAXIMUM, that is, the pattern is bright  

b) to find the distance let's use trigonometry  

   tan θ= y / L  

   tan θ =sin θ/cos θ= sin θ  

because the angles are small  

we substitute  

     y = \frac{m \lambda L}{d} 

let's calculate  

     y = 10 550 10⁻⁹ 1.5 / 1 10⁻³  

     y = 8.25 10⁻³ m

c) for this part the incident light must be white or polychrome light  

let's find the distance (y) for some colors, in the first order  

Violet λ = 400  

y = 1 400 10⁻⁹ 1.5 / 1 10⁻³  

y = 6 10⁻⁴ m  

green λ = 500 nm  

y = 1 500 10⁻⁹ 1.5 / 1 10⁻³  

y = 7.5 10⁻⁴ m  

yellow λ = 600 nm  

y = 1 600 10⁻⁹ 1.5 / 1.10⁻³  

y = 9 10⁻⁴ m  

lam λ = 700 nm  

y = 1 700 10⁻⁹ 1.5 / 1. 10⁻³  

y = 1.05 10⁻³ m  

violet, green, orange, red  

the distance increases with the wavelength increases

d) To find the wavelength we must measure the distance y from the center of the slit for various interference orders and using the equation we can find the wavelength of rationality

4 0
3 years ago
) determine the density of a 32.5 g metal sample that displaces 8.39 ml of water.
sweet-ann [11.9K]
Density is the ratio of a substance's mass to its volume. On the other hand, according to Archimedes' principle, the volume of water displaced is equal to the volume of the object placed on the water. Thus, the density of the metal is equal to 8.39 mL. So, the density would be

Density = 32.5 g/8.39 mL = 3.87 g/mL
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4 years ago
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What mass of water will fill a tank that is 100.0 cm long, 50.0 cm wide, and 30.0 cm high? Express the answer in grams.
Bond [772]
calculate\ mass\ from\ formula\ for\ density:\\\\
density=volume*mass\\
mass=\frac{density}{volume}\\\\
volume=width*length*height\\
volume=100*50*30=150000cm^3=0,15m^3\\\\
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3 years ago
Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a
suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N

Part (b)

the magnitude of each charge, if they have equal magnitude

F = \frac{KQ^2}{r^2}

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q =  \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C

4 0
3 years ago
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