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jolli1 [7]
3 years ago
6

For a cylindrical capacitor, the capacitance does not depend on which of the following values?

Physics
1 answer:
IgorC [24]3 years ago
5 0

Answer:

Capacitance of cylindrical capacitor does not depends on the amount of charge on the conductors

Explanation:

Consider a cylindrical capacitor of length L, inner radius R₁ and outer radius R₂, permitivity ε₀ constant then capacitance of cylindrical capacitor is given by:

C=\frac{2\pi \epsilon_{o}L}{ln\frac{R_{2} }{R_{1}} }

From this equation it is clear that capacitance of cylindrical capacitor is independent of the amount of charge on the conductors where as directly  proportional permitivity constant and length of cylinder where as inversely proportional to natural log of ratio of  R₂ and R₁

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A toaster draws 8 A of current with a voltage of 120 V. Which is the power used by the toaster
Tatiana [17]

Power = (voltage) x (current) =

               (120 V)  x  (8 A)      =  <em>960 watts</em>


7 0
2 years ago
The average speed during any time interval is equal to the total distance of travel divided by the total time. Let d represent t
Tanya [424]

Answer:

Average speed = 3.63 m/s

Explanation:

The average speed during any time interval is equal to the total distance travelled divided by the total time.

That is,

Average speed = distance/ time

Let d represent the distance between A and B.

Let t1 be the time for which she has the higher speed of 5.15 m/s. Therefore,

5.15 = d/t1.

Make d the subject of formula

d = 5.15t1

Let t2 represent the longer time for the return trip at 2.80 m/s . That is,

2.80 = d/t2.

Then the times are t1 = d/5.15 5 and

t2 = d/2.80.

The average speed vavg is given by the following equation.

avg speed = Total distance/Total time

Avg speed = d + d/t1 + t2

Where

Total distance = 2d

Total time = t1 + t2

Total time = d/5.15 + d/2.80

Total time = (2.8d + 5.15d)/14.42

Total time = 7.95d/14.42

Total time = 0.55d

Substitute total distance and time into the formula above.

Avg speed = 2d / 0.55d

Avg Speed = 3.63 m/s

7 0
3 years ago
The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th
Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

E_i = E_f

0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh

v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}

PART B) Replacing the values given as,

h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\

v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}

v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}

v_f = 1.8486m/s

Therefore the speed of the masses would be 1.8486m/s

6 0
3 years ago
A runner drank a lot of water during a race. What is the expected path of the extra filtered water molecules?
Naddika [18.5K]

Answer:

Afferent arteriole, glomerulus, nephron tubule, collecting duct

Explanation:

Blood enters the kidney through the renal artery, a thick branch from the descending aorta. In the hilum, it is divided into several branches that are distributed through the lobes of the kidney and are branching forming numerous afferent arterioles that form the glomerular clew. It is precisely the walls of these capillaries that act as ultrafilters, allowing small particles to pass through.

Blood that flows through the <u>afferent arteriole</u> circulates through the capillary vessels of the kidney (the true capillaries that provide the kidney with oxygen and nutrients necessary for its function). These capillaries are grouped together to form the renal vein which, in turn, pours into the inferior vena cava.

Given the function of the kidneys to eliminate waste products through urine, it is not surprising that these organs are the ones that receive the most blood per gram of weight. One way to express renal blood flow is by considering the renal fraction or fraction of cardiac output that passes through the kidneys.

The regulation of blood flow in the glomeruli is achieved by three formations: the polar bearing, the Goormaghtigh cells and the dense macula. The polar bearing consists of a thickening of the afferent arteriole wall before it enters the <u>renal glomerulus</u>. The arteriole loses its elastic membrane, the endothelium becomes discontinuous and the middle tunic is arranged in two layers, formed by secretory cells: these secretory cells produce Angiotensin and Erythropoietin.

Goormaghtigh cells are arranged at an angle between afferent and effector arterioles and meet in small columns. They are closely related to polar bearing cells. Between both formations is the dense macula (or Zimmerman's dense macula) that is in contact with the distal tubule and afferent arteriole just before it penetrates the glomerulus. These three formations, polar bearing, Goormaghtigh cells and dense macula form the juxtaglomerular apparatus that regulates the blood flow in the glomerulus.

<u>Nephrons</u> regulate water and soluble matter (especially Electrolytes) in the body, by first filtering the blood under pressure, and then reabsorbing some necessary fluid and molecules back into the blood while secreting other unnecessary molecules.

The reabsorption and secretion are achieved with the mechanisms of Cotransporte and Contratransporte established in the nephrons and associated collection ducts. Blood filtration occurs in the glomerulus, a capping of capillaries that is inside a Bowman's capsule.

Liquid flows from the nephron in the <u>collecting duct</u> system. This segment of the nephron is crucial to the process of water conservation by the body. In the presence of the antidiuretic hormone (ADH; also called vasopressin), these ducts become water permeable and facilitate their reabsorption, thus concentrating the urine and reducing its volume. Conversely, when the body must remove excess water, for example after drinking excess fluid, ADH production is decreased and the collecting tubule becomes less permeable to water, making the urine diluted and abundant.

6 0
3 years ago
How much force is required to accelerate a 12 kg mass at 5 m/s 2
Savatey [412]

Answer:

60 N

Explanation:

This is just Newton's Second Law

F = m*a

F = ?

m = 12 kg

a = 5 m/^2

F = 5*12 = 60 Newtons

4 0
2 years ago
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