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3 years ago

Why do two resistors in parallel together contain less resistance than the same two resistors in series?

2 answers:

4 0

Answer : When resistors are connected in parallel, the supply of current is equal to the sum of the currents passing through each of the resistors.

To be specific, the currents in the branches of a parallel circuit add up to the supply current. When resistors are connected in parallel, they have the same potential difference across them.

When resistors are connected in series the total resistance is greater than the individual resistances, hence the currents flow is less.

When resistors are connected in parallel, more current flows from the source than would flow for any of them individual resistors, so the total resistance is lower.

amid [387]3 years ago

4 0

The easiest way to understand that is to think about a 1-lane highway that's
1 mile long. If millions of cars are at one end of it, it can only carry a certain
number of them, and the rest have to wait their turn.

Now build a new highway, exactly the same as the old one ... 1 lane wide and
1 mile long. Put the new one next to the old one. Connect them together at one
end, and connect them together at the other end. Now each car that arrives at
one end has a choice ... take the old road or the new road. In a manner of
speaking, the drivers now encounter less "resistance" in their attempts to get
from here to there ... exactly half as much, in fact ... and twice as many cars
can get through now in the same amount of time.

That's the way I like to explain the idea of parallel resistors. It always helped me
understand them, and I hope it helps you too.

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The United States and South Korean soccer teams are playing in the first round of the World Cup. An American kicks the ball, giv

Aleks04 [339]

Answer:v=2.82 m/s

Explanation:

Given

initial velocity of ball(u)=3.6 m/s

ball rolls a distance(s) of 5 m

ball acceleration $=-0.5 m/s^2$

Velocity of ball when it is intercepted by Korean player

$v^2-u^2=2as$

$v^2=3.6^2+2(-0.5)(5)=7.96$

$v=\sqrt{7.96}$

$v=2.82 m/s$

3 0

3 years ago

Where does the

GREYUIT [131]

Answer:

it comes from your knowledge and the information you have to get the reason why that is the answer so you are putting together things that you already know what the new information you have

5 0

3 years ago

In terms of absorption and reflection, describe why anobject

pishuonlain [190]

Answer & Explanation:

Object are usually considered opaque (its neither transparent or translucent) unless stated otherwise and the color of an opaque object is determine by the color of the light its reflects.

White light comprises of seven constituent of color: red, orange, yellow, green, blue, indigo, violet.

When white light is shown on a red object, it reflects red color only and absorbs all the remaining six colors. Hence, it appears red to the human eye.

4 0

3 years ago

Star a has an absolute magnitude of 8. what is its apparent magnitude at a distance of 100 pc

tresset_1 [31]

From the definition of apparent magnitude, we know that:
$m_{1} - m_{2} = -2.5Log( \frac{F_{1}}{F_{2}} )$
where:
m = apparent magnitude
F = corresponding flux

We also know that the flux is given by the formula:
$F = \frac{L}{4 \pi d^{2} }$
where:
L = luminosity
d = distance

Therefore:
$\frac{F_{1} }{F_{2}} = (\frac{L_{1} }{4 \pi d_{1}^{2} })(\frac{4 \pi d_{2}^{2} }{L_{2} }) \\ = \frac{L_{1}d_{2}^{2}}{L_{2}d_{1}^{2}}$

Now, let's apply these formulae to the same star (therefore, same luminosity), using apparent magnitude and absolute magnitude (which is defined as the apparent magnitude the star would have if it were at a distance of 10pc):
$m - M = 2.5 Log ( \frac{d}{10pc})^{2}$

Now, let's solve for m:
$m = M + 2.5 Log ( \frac{d}{10})^{2}$
= <span> $8 + 2.5 Log ( \frac{100}{10})^{2}$ </span>
= 13

Hence, the apparent magnitude of the star would be m = +13

3 0

3 years ago

You are at the park with your little brother, when you notice a small merry-go-round with a radius that looks to be about 1.5 m.

Elodia [21]

Answer:

Explanation:

angle covered in one rotation = 2π radian

θ = ωt + 1/2 αt²

θ is angle rotated in time t with initial angular velocity of ω and angular acceleration α .

Putting the values

2π = 0 + 1/2 x α x 3²

α = 1. 4 radian / s²

linear acceleration = α x r = 1.4 x 1.5 = 2.1 m / s².

Initial acceleration = 2.1 m /s²

final angular velocity = α t = 1.4 x 3 = 4.2 radian / s

linear velocity = 4.2 x 1.5 = 6.3 m /s

centripetal acceleration = v² / R = 6.3² / 1.5 = 26.46 m /s²

radian acceleration = 26.46 m /s

tangential acceleration = 2.1 m /s²

Total final acceleration = √ ( 26.46² + 2.1² )

= √ ( 700.13 + 4.41)

Final acceleration = 26.53 m / s²

7 0

3 years ago