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Oksanka [162]
3 years ago
11

What is the equivalent resistance in this circuit? What is the current in this circuit?

Physics
2 answers:
suter [353]3 years ago
5 0

Answer:

15 Ohms

3 Amps

Explanation:

kramer3 years ago
3 0

All these resistors are in series so we can take the sum of them by:

Rtotal = R1 + R2 + R3......

So...

Rtotal = 2 + 3 + 4 + 6

Rtotal = 15

So now the total resistance in the circuit is 15 ohms and the potential difference applied to the circuit is 45 volts

Now we can use:

V = IR

Isolate for I

V/R = I

45/15 = I

I = 3 amps (A)

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Explanation:

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The charge per unit length of the wire is \lambda and the net charge per unit length is 2 \lambda.

We know that there exist zero electric field inside the metal cylinder.

(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let \lambda_i\ and\ \lambda_o are the charge per unit length on the inner and outer surfaces of the cylinder.

For inner surface,

\phi=\dfrac{q_{enclosed}}{\epsilon_o}

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\lambda_i=-\lambda  

For outer surface,

\lambda_i+\lambda_o=2\lambda

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\lambda_o=3\lambda

(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :

E_o=\dfrac{\lambda_o}{2\pi \epsilon_o r}

E_o=\dfrac{3\lambda}{2\pi \epsilon_o r}

Hence, this is the required solution.

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3 years ago
On a planet where g = 10.0 m/s2 and air resistance is negligible, a sled is at rest on a rough inclined hill rising at 30°. The
Cloud [144]

Answer:

Explanation:

a is the acceleration

μ is the coefficient of friction

Acceleration of the object is given by

a = g (\sin  \theta -\mu \cos \theta)\\\\=10( \sin 30 - 0.4 \cos 30)\\\\=10(0.5-0.3464)\\\\=1.54m/s^2

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after travelling 4m , its velocity becomes 0

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a=\frac{0-u^2}{2s}

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μ = F/N

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On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-f
aliya0001 [1]

Answer:

15.3 s and 332 m

Explanation:

With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon

    gm = 1/6 ge

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We calculate the range

    R = Vo² sin 2θ  / g

    R = 25² sin (2 30) / 1.63

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We will calculate the time of flight,

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0 = Vo sin  t – ½  g t2

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0= 12.5 t –  0.815 t2

We solve the equation

0= t ( 12.5 -0.815 t)

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t= 15.3 s

The value of zero corresponds to the departure point and the flight time is 15.3 s

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R2 = 25² sin (2 30) / 9.8

R2 = 55.2 m

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R/R2 = 6

Therefore the ball travels a distance six times greater on the moon than on Earth

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<span>2nd Law: Force is equal to the mass times acceleration. </span>

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hope this helps and if you have any questions just hmu and ask :)</span>
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3 years ago
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