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bulgar [2K]
3 years ago
9

Consider the circuit shown in the figure below. (Let R1 = 3.00 Ω, R2 = 8.00 Ω, and = 10.0 V.). . (a) Find the voltage across R1.

. . . . . (b) Find the current in R1..
Physics
1 answer:
Goryan [66]3 years ago
8 0
You take the inverse of the total resistances of each branch and add them up. So if you have 5ohm, 7 ohm, and 10ohm, you would add 1/5 + 1/7 + 1/10 = 31/70 Then flip it back by either using the <span>x<span>−1</span></span><span> (inverse) key on your calculator or simply dividing 70 by 31 to get a total of 2.26ohms</span>
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Answer:

The gravitational force on the elevator = 4500N

Explanation:

The given parameters are;

The force applied by the elevator, F  = 4500 N

The acceleration of the elevator = Not accelerating

From Newton's third law of motion, the action of the cable force is equal to the reaction of the gravitational force on the elevator which is the weight, W and motion of the elevator as follows;

F = W + Mass of elevator × Acceleration of elevator

∴ F = W + Mass of elevator × 0 = W

F = 4500 N = W

The net force on the elevator is F - W = 0

The gravitational force on the elevator = W = 4500N.

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Answer

Time period T = 1.50 s

time t = 40 s

r = 6.2 m

a)

Angular speed ω = 2π/T

                              = \dfrac{2\pi }{1.5}  

                              = 4.189 rad/s

Angular acceleration α = \dfrac{\omega}{t}

                                      = \dfrac{4.189}{40}

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Tangential acceleration a = r α = 6.2 x 0.105 = 0.651 m/s²

b)The maximum speed.

       v = 2πr/T

          = \dfrac{2\pi \times 6.2}{1.5}

          = 25.97 m/s

So centripetal acceleration.

        a = \dfrac{v^2}{r}

          = \dfrac{25.97^2}{6.2}

          =  108.781 m/s^2

          = 11.1 g    

in combination with the gravitation acceleration.

a_{total} = \sqrt{(11.1g)^2+g^2}

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