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3 years ago

9

Discuss the advantages and disadvantages of series circuits compared to parallel circuits. Give an example of where each type of

circuit could be used.

1 answer:

Murljashka [212]3 years ago

5 0

Advantage of series circuit :

1. allows all the components in the circuit to have same current flowing through them.

2. one switch is enough to turn off all the components in the circuit in series.

3. make the circuit small by using smaller length of wire.

Disadvantages of series circuit :

1. If one component becomes defective , it does not allow other components to operate, since they share same current.

2. all components can not have same voltage across them.

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A cannon, elevated at 40∘ is fired at a wall 300 m away on level ground, as shown in the figure below. The initial speed of the

Gre4nikov [31]

Vo = 89 m/s
angle: 40°

=> Vox = Vo * cos 40° = 89 * cos 40°

=> Voy = Vo. sin 40° = 89 * sin 40°

x-movement: uniform => x =Vox * t = 89*cos(40)*t

x = 300 m => t = 300m / [89m/s*cos(40) = 4.4 s

y-movement: uniformly accelerated => y = Voy * t - g*t^2 /2

y = 89m/s * sin(40) * (4.4s) - 9.m/s^2 * (4.4)^2 / 2 = 156.9 m = height the ball hits the wall.

7 0

3 years ago

Read 2 more answers

What potential difference is required in an electron microscope to give an electron wavelength of 4. 5 nm?

Lorico [155]

Potential difference required in an electron microscope to give an electron wavelength of 4. 5 nm will be 0.063 V.

The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other is called potential difference.

The wavelength of an electron is calculated for a given energy (accelerating voltage) by using the de Broglie relation between the momentum p and the wavelength λ of an electron

lambda = 4.5 nm = 4.5 * $10^{-9}$ m

h = $6.626 * 10^{-34}$ J s

e = 1.6 * $10^{-19}$ C

m = 9.1 * $10^{-31}$ kg

Energy = eV

lambda = h / $\sqrt{2mE}$ = h / $\sqrt{2m(eV)}$

$(lambda)^{2}$ = $h^{2}$ / (2m (eV))

V = $h^{2}$ / (2 m e $(lambda)^{2}$ )

V = $(6.626 * 10^{-34} )^{2}$ / 2 * 9.1 * $10^{-31}$ * 1.6 * $10^{-19}$ * $(4.9 * 10^{-9}) ^{2}$

V = 0.063 V

To learn more about wavelength of an electron here

brainly.com/question/17295250

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2 years ago

A plane travels at a speed of 205mph in still air. Flying with a tailwind, the plane is clocked over a distance of 1000 miles. F

vaieri [72.5K]

While plane is moving under tailwind condition it took time "t"

so here we will have

$t = \frac{d}{v_{net}}$

here net speed of the plane will be given as

$v_{net} = v + v_w$

$t = \frac{1000}{205 + v_w}$

similarly when it moves under the condition of headwind its net speed is given as

$v_{net} = v - v_w$

now time taken to cover the distance is 2 hours more

$t + 2 = \frac{1000}{205 - v_w}$

now solving two equations

$\frac{1000}{205 + v_w} + 2 = \frac{1000}{205 - v_w}$

solving above for v_w we got

$v_w = 40.4 mph$

6 0

3 years ago

A descent vehicle landing on the moon has a vertical velocity toward the surface of the moon of 29.6 m/s. At the same time, it h

inessss [21]

If the vertical component is 29.6 m/s down, and the horizontal component
is 54.8 m/s parallel to the surface, then the magnitude of the slanty vector is

√(29.6² + 54.8²) = √(876.16 + 3003.04) = √3879.2 = 62.28 m/s .

That's 139 mph ! Wow !

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3 years ago

Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground. Think about the amounts o

likoan [24]

Answer:

1) At the highest point of the building.

2) The same amount of energy.

3) The kinetic energy is the greatest.

4) Potential energy = 784.8[J]

5) True

Explanation:

Question 1

The moment when it has more potential energy is when the ball is at the highest point in the building, that is when the ball is at a height of 40 meters from the ground. It is taken as a point of reference of potential energy, the level of the soil, at this point of reference the potential energy is zero.

$E_{p} = m*g*h\\E_{p} = 2*9.81*40\\E_{p} = 784.8[J]$

Question 2)

The potential energy as the ball falls becomes kinetic energy, in order to be able to check this question we can calculate both energies with the input data.

$E_{p}=m*g*h\\ E_{p} = 2*9.81*20\\ E_{p} = 392.4[J]\\$

And the kinetic energy will be:

$E_{k}=0.5*m*v^{2}\\ where:\\v = velocity = 19.8[m/s]\\E_{k}=0.5*2*(19.8)^{2}\\ E_{k}=392.04[J]$

Therefore it is the ball has the same potential energy and kinetic energy as it is half way through its fall.

Question 3)

As the ball drops all potential energy is transformed into kinetic energy, therefore being close to the ground, the ball will have its maximum kinetic energy.

$E_{k}=E_{p}=m*g*h = 2*9.81*40\\ E_{k} = 784.8[J]\\ E_{k} = 0.5*2*(28)^{2}\\ E_{k} = 784 [J]$

Question 4)

It can be easily calculated using the following equation

$E_{p} =m*g*h\\E_{p}=2*9.81*40\\E_{p} =784.8[J]$

Question 5)

True

The potential energy at 20[m] is:

$E_{p}=2*9.81*20\\ E_{p}= 392.4[J]\\The kinetic energy is:\\E_{k}=0.5*2*(19.8)^{2} \\E_{k}=392[J]$

3 0

3 years ago