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Wittaler [7]
4 years ago
10

The maximum speed around a level curve is 30.0 km/h. What is the maximum speed around a curve with twice the radius?

Physics
2 answers:
erastovalidia [21]4 years ago
6 0

Answer:

When the radius is doubled, its speed will be 42.42 km/h.

Explanation:

Given that,

The maximum speed around a level curve is 30 km/h = 8.34 m/s

We need to find the maximum speed around a curve with twice the radius. The maximum speed of the object is given by :

v=\sqrt{\mu rg}............(1)

If radius is doubled, r' = 2r

The maximum speed is given by :

v'=\sqrt{\mu r'g}

v'=\sqrt{\mu (2r)g}..........(2)

Dividing equation (1) and (2) we get :

\dfrac{v}{v'}=\dfrac{1}{\sqrt2}

v'=\sqrt{2} v

v'=\sqrt{2}\times 30

v' = 42.42 km/h

So, when the radius is doubled, its speed will be 42.42 km/h. Hence, this is the required solution.

Evgen [1.6K]4 years ago
3 0

Answer:

The maximum speed around a curve is 42.4 km/h.

Explanation:

Given that,

Maximum speed = 30.0 km/h

New radius = 2 r

The centripetal force will be same for both situation.

We need to calculate the new maximum speed around a curve

Using formula of  centripetal force

\dfrac{mv^2}{r}=\dfrac{mv^2}{r}

Put the value into the formula

\dfrac{v^2}{2r}=\dfrac{30^2}{r}

v=30\sqrt{2}\ km/h

v=42.4\ km/h

Hence, The maximum speed around a curve is 42.4 km/h.

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  • The total current in the circuit is 0.5 Ampere.
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  • The voltage drop atR_1 is 4 volts.
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R_{eq}=R_1+R_2+R_3\\=8 \Omega +5 \Omega + 3\Omega =16\omega

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V=IR_{eq}\\I=\frac{V}{R_{eq}}=\frac{8 V}{16 \Omega}=0.5 A (Ohm's law)

  • For series combinations, the current in each resistor remains the same.

So, the current in R_1, R_2 \&R_3:

I_1= I_2= I_3=I=0.5 A\\

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The current across  R_1 = I = 0.5 A

V_1=I\times R_1\\\\=0.5A\times 8\Omega = 4 V

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The current across  R_2 = I = 0.5 A

V_2=I\times R_2\\\\=0.5A\times 5\Omega = 2.5 V

  • The voltage drop across at R_3 = V_3

The current across  R_3 = I = 0.5 A

V_3=I\times R_3\\\\=0.5A\times 3\Omega = 1.5 V

  • The total power consumed by circuit:

P= V\times I \\\\= 0.5 A\times 8 V = 4 watt

  • Power consumed at R_1:

P_1=V_1\times I\\\\= 4V\times 0.5A = 2 watt

  • Power consumed at R_2:

P_2=V_2\times I\\\\= 2.5 V\times 0.5A = 1.25 watt

  • Power consumed at R_3:

P_3=V_3\times I= \\\\1.5 V\times 0.5A = 0.75 watt

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