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Katyanochek1 [597]
3 years ago
14

10Ω resistor and a 5 Ω are connected in parallel. This pair is then connected in series in with another 4 Ω resistor. What is th

e equivalent resistance of the circuit?
Physics
1 answer:
Dmitrij [34]3 years ago
5 0

Answer:

7.33 Ω

Explanation:

10Ω resistor and a 5 Ω are connected in parallel

so,

R₁=(10X5)/(10+5)

   =(50/15) Ω

   =3.33 Ω

This pair is then connected in series in with another 4 Ω resistor

R₂=R₁+4

    =(3.33+4) Ω

    =7.33 Ω

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Answer:

(a). The acceleration is 3.20 m/s².

(b). The amount of work done by each force is 19.2 J.

(c). The total work done on the block is 19.2 J.

(d). The final speed of the block is 6.26 m/s.

Explanation:

Given that,

Mass of block = 3 kg

Distance = 2 m

Angle = 30°

Coefficient of kinetic friction = 0.20

(a). We need to calculate the acceleration

Using balance equation of force

N = mg\co\theta

mg\sin\theta-f_{k}=ma

a = g\sin\theta-\mu g\cos30

Put the value into the formula

a=g(\sin30-0.20\cos30)

a=9.8(\dfrac{1}{2}-0.20\times\dfrac{\sqrt{3}}{2})

a=3.20\ m/s^2

The acceleration is 3.20 m/s².

(b). We need to calculate the amount of work done by each force

Using formula of work done

Normal force is

N = mg\cos30

So due to this the net force is zero then the no work done by reaction force.

By another force,

W= F\times d

W=ma\times d

Put the value into the formula

W= 3\times3.20\times2

W=19.2\ J

The amount of work done by each force is 19.2 J.

(c). We need to calculate the total work done on the block

The total work done on the block is 19.2 J.

(d). We need to calculate the final speed of the block

Using equation of motion

v^2=u^2+2gs

Put the value into the formula

v^2=0+2\times9.8\times2

v=\sqrt{2\times9.8\times2}

v=6.26\ m/s

The final speed of the block is 6.26 m/s.

Hence, This is the required solution.

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The moment of inertia of a thin ring of mass M and radius R about its symmetry axis is ICM = MR2.
monitta

Answer:

The moment of inertia of large ring is 2MR².

(A) is correct option.

Explanation:

Given that,

Mass of ring = M

Radius of ring = R

Moment of inertia of a thin ring = MR²

Moment of inertia :

Moment of inertia is the product of the mass of the ring and square of radius of the ring.

We need to calculate the moment of inertia of large ring

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Put the value into the formula

I=MR^2+MR^2

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Hence, The moment of inertia of large ring is 2MR².

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