All categories

2 years ago

What is the effective resistance of this dc circuit

Physics

1 answer:

Inessa05 [86]2 years ago

5 0

You've managed somehow to post the mirror image of the circuit diagram, including the numbers and values of the resistors. I'm curious to know how you did that.

The three resistors at the left end of the diagram are 3Ω , 2Ω , and 1Ω all in series. They behave like a single resistor of (3+2+1) = 6Ω .

That 6Ω resistor is in parallel with the 2Ω drawn vertically in the middle of the diagram. That combination acts like a single resistor of 1.5Ω in that position.

Finally, we have that 1.5Ω resistor in series with 1Ω and 4Ω . That series combination behaves like a single resistor of <em>6.5Ω</em> across the battery V.

You might be interested in

The core of a star must be at temperature of _____ degrees Celsius for hydrogen fusion to take place. 10,000 100,000 1,000,000 1

disa [49]

The core of a star must be at the temperature of 10,000,000 degrees Celsius for hydrogen fusion to begin.

6 0

3 years ago

The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a

likoan [24]

Answer:

14869817.395 m

Explanation:

$\theta$ =22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

$22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians$

From Rayleigh Criterion

$\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m$

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0

3 years ago

If you had started with a larger mass, how would the half-life change?

iogann1982 [59]

Answer:

There is no change, unless your mass is somehow at the quantum level, at which the concept of half-life breaks down.

Half life is a property of the specific radioactive isotope...NOT of the initial sample's mass.

3 0

2 years ago

The latitude of the city of Arlington is about 32.7357°. Calculate the following: (a) The angular speed of the Earth. (b) The li

julia-pushkina [17]

Answer:

Explanation:

a ) The earth rotates by 2π radian in 24 x 60 x 60 s

so angular speed ( w ) = 2π / (24 x 60 x 60) = 7.268 x 10⁻⁵ rad / s

b ) Linear speed of city of Arlington ( v ) = w r = w R Cosλ where R is radius of the earth and λ is latitude .

v = 7.268 x 10⁻⁵ x 6.371 x 10⁶ cos 32.7357

389.5 m /s

acceleration = w² r = w² R Cos 32.7357

= (7.268 x 10⁻⁵ )² x 6.371 x 10⁶ x cos 32.7357

=283.08 x 10⁻⁴ m/s²

c) velocity ratio =

w r /w R =

R cos 32.73/ R

= Cos 32.73

= 0.84 .

6 0

3 years ago

A football player with a mass of 88 kg and a speed of 2.0 m/s collides head-on with a player from the opposing team whose mass i

Ket [755]

Answer:

Speed of another player, v₂ = 1.47 m/s

Explanation:

It is given that,

Mass of football player, m₁ = 88 kg

Speed of player, v₁ = 2 m/s

Mass of player of opposing team, m₂ = 120 kg

The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :

$m_1v_1+m_2v_2=(m_1+m_2)V$

V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.

$v_2=-\dfrac{m_1v_1}{m_2}$

$v_2=-\dfrac{88\ kg\times 2\ m/s}{120\ kg}$

$v_2=-1.47\ m/s$

So, the speed of another player is 1.47 m/s. Hence, this is the required solution.

7 0

3 years ago