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77julia77 [94]
3 years ago
11

An oceanographer is studying how the ion concentration in seawater depends on depth. She makes a measurement by lowering into th

e water a pair of concentric metallic cylinders at the end of a cable and taking data to determine the resistance between these electrodes as a function of depth. The water between the two cylinders forms a cylindrical shell of inner radius r_a, outer radius r_b, and length L much larger than r_b. The scientist applies a potential difference Delta V between the inner and outer surfaces, producing an outward radial current I. Let rho represent the resistivity of the water.
(a) Find the resistance of the water between the cylinders in terms of L, rho, r_a, and r_b.
(b) Express the resistivity of the water in terms of the measured quantities L, r_a, r_b, Delta V and I.
Physics
1 answer:
Black_prince [1.1K]3 years ago
6 0

Answer:

a)  R = ρ₀ L /π(r_b² - R_a²) , b)  ρ₀ = V / I    π (r_b² - R_a²) / L

Explanation:

a) The resistance of a material is given by

          R = ρ l / A

where ρ is the resistivity, l is the length and A is the area

the length is l = L and the resistivity is ρ = ρ₀

the area is the area of ​​the cylindrical shell

           A = π r_b² - π r_a²

           A = π (r_b² - r_a²)

we substitute

         R = ρ₀ L /π(r_b² - R_a²)

b) The potential difference is related to current and resistance by ohm's law

         V = i R

         

we subsist the expression of resistance

          V = I ρ₀ L /π (r_b² - R_a²)

           ρ₀ = V / I    π (r_b² - R_a²) / L

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You, a 70 kg person, leap from a 10 m tall building and land feet first on a trampoline. The center of the trampoline where you
Anon25 [30]

Answer:

1373.4 N/m

Explanation:

Hooke's law states that the extension of a spring and force are related by the expression, F=kx where k is spring constant, x is extension of spring and F is the applied force. Making k the subject of the formula then

k=\frac {F}{x}

Also, F=gm hence the above formula is modified as

k=\frac {gm}{x}

Taking g as 9.81 m/s2 , x as 0.5 m and m as 70 kg then

k=\frac {9.81\times 70 kg}{0.5m}=1373.4 N/m

4 0
3 years ago
n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.
9966 [12]

Answer:

a). V = 3.13*10⁶ m/s

b). T = 1.19*10^-7s

c). K.E = 2.04*10⁵

d). V = 1.02*10⁵V

Explanation:

q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

B = 1.10T

1u = 1.67 * 10^-27kg

M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

a). Centripetal force = magnetic force

Mv / r = qB

V = qBr / m

V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

V = 2.09088 * 10^-20 / 6.68 * 10^-27

V = 3.13*10⁶ m/s

b). Period of revolution.

T = 2Πr / v

T = (2*π*0.0594) / 3.13*10⁶

T = 1.19*10⁻⁷s

c). kinetic energy = ½mv²

K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²

K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

V = K / q

V = 2.04*10⁵ / (2eV).....2e-

V = 1.02*10⁵V

7 0
3 years ago
A 615.00 kg race car is uniformly traveling around a circular race track. It takes the race car 20.00 seconds to do one lap arou
koban [17]

Answer:

The value is  f  =  0.05 \ Hz

Explanation:

From the question we are told that

     The mass of the car is  m  =  615 \  kg

      The period of the circular motion is  T  =  20 \  s

      The radius  is r =  80 \  m/s

Generally the frequency of the circular motion is  

       f  =   \frac{1}{T }

=>    f  =   \frac{1}{ 20  }

=>    f  =  0.05 \ Hz

3 0
3 years ago
The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1
telo118 [61]
<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life h of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: h_{B-13}=5(10)^{-10}s=0.0000000005s

beryllium-15: h_{B-15}=2(10)^{-7}s=0.0000002s

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

h_{B-15}=X.h_{B-13}

Where X is the amount we want to find:

X=\frac{h_{B-15}}{h_{B-13}}

X=\frac{2(10)^{-7}s}{5(10)^{-10}s}

Finally:

X=400

Therefore:

The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.

8 0
3 years ago
What is the difference in Neil Armstrong’s weight on the moon and on earth? Neils mass is 160kg including his spacesuit and back
Len [333]

Explanation:

Given parameters:

Mass of Neil Armstrong = 160kg

Gravitational pull of earth = 10N/kg

Moon's pull = 17% of the earth's pull

Unknown:

Difference between Armstrong's weight on moon and on earth.

Solution:

To find the weight,

   Weight = mass x acceleration due to gravity = mg

Moon's gravitational pull = 17% of the earth's pull = 17% x 10 = 1.7N/kg

Weight on moon = 160 x 1.7 = 272N

Weight on earth = 160 x 10 = 1600N

The difference in weight = 1600 - 272 = 1328N

The weight of Armstrong on earth is 1328N more than on the moon.

Learn more:

Weight and mass brainly.com/question/5956881

#learnwithBrainly

3 0
3 years ago
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