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2 years ago

Number of conducting plates of a multiplate capacitor is 5. The no. Of capacitors is

Physics

1 answer:

Ivahew [28]2 years ago

8 0

Answer:

4 capacitors

Explanation:

Given

$n = 5$ --- conducting plates

Required

The number of capacitor (c)

This is calculated as:

$c = n - 1$

So, we have:

$c = 5 - 1$

$c = 4$

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Write an email to a classmate explaining why the velocity of the current in a river has no FN C02-F20-OP11USB effect on the time

anzhelika [568]

The velocity of the current in a river has no effect on the the time it takes to paddle a canoe across the river, given that the boat is pointed perpendicular to the bank of the river, because

The velocity of the river does not change the velocity and therefore the distance traveled in the direction of the boat which is directed perpendicular to it

Reason:

Let $\overset \rightarrow {v_y}$ represent the velocity of the boat across the river in the direction, $\overset \rightarrow {d_y}$ , and let, $\overset \rightarrow {v_x}$ , represent the velocity of the river, we have;

The velocity of the boat perpendicular to the direction of the river = $\overset \rightarrow {v_y}$

Therefore, the distance covered per unit time in the perpendicular direction

to the flow of the river is $\overset \rightarrow {v_y}$ , such that the time it takes to cross the river in

the perpendicular direction is the same, for every value of the velocity of

the river.

This is so because the velocity in the perpendicular direction to the flow of

the river, which is the velocity of the boat is unchanged by the velocity of

the river, because there is no perpendicular component of velocity in the

velocity of the river.

$\overset \rightarrow {v_y}$ = 3 m/s

$\overset \rightarrow {v_x}$ = 4 m/s

The width of the river, $w_y$ = 6 meters, we have;

The resultant velocity = $\sqrt{(3 \ m/s)^2 + (4 \ m/s)^2} =5 \ m/s$

The direction, θ, is given as follows;

$\theta = \arctan \left(\dfrac{4}{3} \right) \approx 53.13^{\circ}$

The length of the path of the boat, <em>l</em>, is given as follows;

$l = \dfrac{6}{cos \left(\arctan \left(\dfrac{4}{3} \right)\right)} = 10$

The length of the path the boat takes = 10 m

The time it takes to cross the river, t = $\dfrac{l}{v}$ , therefore;

$t = \dfrac{10}{5} = 2$

The time it takes to cross the river, t = 2 seconds

Considering only the y-components, we have;

$t = \dfrac{w_y}{v_y}$

Therefore;

$t = \dfrac{6 \ m}{3 \ m/s} = 2 \, s$

Which expresses that the time taken is the same and given that the

vectors of the velocities of the river and the boat are perpendicular, the

distance covered in the direction of the boat is unaffected by the velocity

of the river.

Learn more vectors here:

brainly.com/question/15907242

5 0

3 years ago

If the _____ work on a simple machine equals the output work, the machine is said to be 100% efficient.

omeli [17]

<h3>Answer:</h3>

Input work

<h3>Explanation:</h3>

Concept being tested: Efficiency of machines

Therefore we need to know what is the efficiency of a machine

Efficiency of a machine is the ratio of work output of machine to the work input expressed as a percentage.

Thus;

Efficiency = (Work output ÷ Work input) × 100%

Therefore, if the work input is equal to the work output then the efficiency of the machine will be 100%.

Most machines are not 100% efficient due to loss of energy in form of heat due to friction of the moving parts of the machine.

3 0

3 years ago

Archaeologists use the radioactive decay of carbon-14 to estimate the age of fossils and can be used on samples of bone, cloth,

timama [110]

The answer should be 11,460 because the first half-life leaves 50 percent left and the next half-life would leave 25 percent which dates the bones at 11,460 years old.

8 0

2 years ago

Read 2 more answers

How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the

kirza4 [7]

Answer:

Approximately $6.2\; {\rm rpm}$ , assuming that the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ .

Explanation:

Let $\omega$ denote the required angular velocity of this Ferris wheel. Let $m$ denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

Weight of the passenger (downwards), $m\, g$ , and possibly
Normal force $F_\text{normal}$ that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is $0$ - that is, $F_\text{normal} = 0$ .

The net force on this passenger is $(m\, g - F_\text{normal})$ . Hence, when $F_\text{normal} = 0$ , the net force on this passenger would be equal to $m\, g$ .

Passengers on this Ferris wheel are in a centripetal motion of angular velocity $\omega$ around a circle of radius $r$ . Thus, the centripetal acceleration of these passengers would be $a = \omega^{2}\, r$ . The net force on a passenger of mass $m$ would be $m\, a = m\, \omega^{2}\, r$ .

Notice that $m\, \omega^{2} \, r = (\text{Net Force}) = m\, g$ . Solve this equation for $\omega$ , the angular speed of this Ferris wheel. Since $g = 9.81\; {\rm m\cdot s^{-2}}$ and $r = 23\; {\rm m}$ :

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$ .

$\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}$ .

The question is asking for the angular velocity of this Ferris wheel in the unit ${\rm rpm}$ , where $1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})$ . Apply unit conversion:

$\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}$ .

3 0

1 year ago

A solid cylinder with a mass of 2.46 kg and a radius of 0.049 m starts from rest at a height of 4.80 m and rolls down a 24.3 ◦ s

irina1246 [14]

Answer:

$v_{f}\approx 2.097\,\frac{m}{s}$

Explanation:

Let assume that the solid cylinder rolls down a frictionless incline. The translational speed can be found by using the Principle of Energy Conservation and the Work-Energy Theorem:

$m_{cyl}\cdot g\cdot y_{o} = \frac{1}{2}\cdot m_{cyl} \left( 1+ \frac{1}{R}\right)\cdot v_{f}^{2}$

$g\cdot y_{o} = \frac{1}{2}\cdot\left( 1+ \frac{1}{R}\right)\cdot v_{f}^{2}$

The translational speed is:

$v_{f} = \sqrt{\frac{2\cdot g\cdot y_{o}}{\left(1 + \frac{1}{R} \right)}}$

$v_{f} = \sqrt{\frac{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (4.80\,m)}{\left(1 + \frac{1}{0.049\,m} \right)} }$

$v_{f}\approx 2.097\,\frac{m}{s}$

7 0

3 years ago