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Lina20 [59]
3 years ago
15

A 16.0 Ω, 13.0 Ω, and 7.00 Ω resistor are connected in parallel to an emf source. A current of 6.00 A is in the 13.0 Ω resistor.

Calculate the equivalent resistance of the circuit. Answer in units of Ω.
Physics
2 answers:
Firlakuza [10]3 years ago
4 0

The current in one parallel resistor doesn't tell us the equivalent resistance of the circuit. And the equivalent resistance of the circuit doesn't depend on the current through any of the resistors, or the voltage applied to them, or even whether there IS any current through any of them.  

Resistors in parallel have the same equivalent resistance even if they're wrapped in tissue, placed in a matchbox, and stored in the back of a drawer in a desk and locked in a dusty warehouse.

The equivalent resistance of 16Ω, 13Ω, and 7Ω all in parallel is the reciprocal of

(1/16) + (1/13) + (1/7) .

That's  1 / (0.0625 + 0.0769 + 0.1429)

That's  1 / (0.2823)

That's   <em>3.543 Ω</em> .

Darya [45]3 years ago
3 0

Answer:3.54ohms

Explanation: connection in parallel

1/Rt= 1/R1+1/R2+1/R3

1/Rt= 1/16+1/13+1/7

1/Rt= 91+112+208/1456

1/Rt= 411/1456

411Rt= 1456

Rt= 1456/411

Rt= 3.54ohms

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A new planet has been discovered and given the name Planet X . The mass of Planet X is estimated to be one-half that of Earth, a
harina [27]

Answer:

    vₐ = v_c  ( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )

Explanation:

To calculate the escape velocity let's use the conservation of energy

starting point. On the surface of the planet

          Em₀ = K + U = ½ m v_c² - G Mm / R

final point. At a very distant point

         Em_f = U = - G Mm / R₂

energy is conserved

           Em₀ = Em_f

           ½ m v_c² - G Mm / R = - G Mm / R₂

           v_c² = 2 G M (1 /R -  1 /R₂)

if we consider the speed so that it reaches an infinite position R₂ = ∞

           v_c = \sqrt{\frac{2GM}{R} }

now indicates that the mass and radius of the planet changes slightly

            M ’= M + ΔM = M ( 1+ \frac{\Delta M}{M} )

            R ’= R + ΔR = R ( 1 + \frac{\Delta R}{R} )

we substitute

           vₐ = \sqrt{\frac{2GM}{R} } \  \frac{\sqrt{1+ \frac{\Delta M}{M} } }{ \sqrt{1+ \frac{ \Delta R}{R} } }

         

let's use a serial expansion

           √(1 ±x) = 1 ± ½ x +…

we substitute

         vₐ = v_ c ( (1 + \frac{1}{2}  \frac{\Delta M}{M} )  \ ( 1 - \frac{1}{2}  \frac{\Delta R}{R} ))

we make the product and keep the terms linear

        vₐ = v_c  ( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )

5 0
2 years ago
So I'm struggling with rearranging kinematic formulas. Does anyone have any steps or something to help.
bekas [8.4K]

Rearranging formulas is all about simple algebra rules. Just like when solving for x in an equation, you're just isolating whichever variable you want. I'll work this one out for you and hopefully it'll help, but if you need more explanation, then feel free to comment!

D = ViT + 0.5at²   Subtract ViT from both sides

D - ViT = 0.5at²    Divide both sides by 0.5t²

\frac{D - ViT}{0.5t^{2} } = \frac{0.5at^{2} }{0.5t^{2} }    I wrote this step out a little more to show how your fraction will cancel

\frac{D - ViT}{0.5t^{2} }= a    I like to flip these around so the single variable is on the right

a = \frac{D - ViT}{0.5t^{2} }

7 0
3 years ago
PLEASE HELP!!!!!!! MT TIME FOR MY TEST IS ALMOST OVER!!!
pav-90 [236]

Answer:

50 N

Explanation:

Let the natural length of the spring = L

so

100 = k(40 - L)       (1)

200 = k(60 - L)       (2)

(2)/(1):   2 = (60 - L)/(40 - L)

60 - L = 2(40 - L)

60 - L = 80 - 2L

2L - L = 80 - 60

L = 20

Sub it into (1):

100 = k(40 - 20) = 20k

k = 100/20 = 5 N/in

Now

X = k(30 - L) = 5(30 - 20) = 50 N

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