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alisha [4.7K]
2 years ago
11

8. Calculate the total resistance in a series circuit with a fan (3 ohms) and 3 lights (1 ohms each). 9 ohms O 12 ohms 15 ohms 0

6 ohms​
Physics
2 answers:
Paraphin [41]2 years ago
5 0

Answer:

 R_eq = 6 Ω

Explanation:

In a series circuit the current is constant throughout the circuit and the resistance is the sum of all the resistances in it.

         R_eq = ∑

in this case we have

         R_eq = R₁1 + 3 R₂

where R₁ is the resistance of the fan and R₂ the resistance of each light, we must clarify that we will assume that the resistance of the lights does not change with temperature

let's calculate

          R_eq = 3 + 3 1

           R_eq = 6 Ω

aivan3 [116]2 years ago
5 0

Answer:

9 ohms

Explanation:

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What is the mechanical advantage of the machine shown below? a 5 b 4 c 3 d 2
Alex

Answer:

Explanation: Mechanical advantage is a measure of the force amplification achieved by using a tool, mechanical device or machine system. The device preserves the input power and simply trades off forces against movement to obtain a desired amplification in the output force.

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2 years ago
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

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A boy whirls a stone in a horizontal circle of radius 1.4 m and at height 1.5 m above ground level. The string breaks, and the s
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Explanation:

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R=u\times t

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u=18.07 m/s

Therefore magnitude of centripetal acceleration is given by

a_c=\frac{u^2}{r}=\frac{18.07^2}{1.4}=233.23 m/s^2

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