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Leto [7]
2 years ago
8

The maximum centripetal acceleration a car can sustain without skidding out of a curved path is 9.40 m/s2 . If the car is travel

ing at a constant speed of 40.0 m/s on level ground, what is the radius R of the tightest unbanked curve it can negotiate?
Physics
1 answer:
AnnyKZ [126]2 years ago
8 0

Answer:

The radius = 170.21 m

Explanation:

The given data are : -

The centripetal acceleration of a car = 9.40 m/s².

Speed of a car = 40.0 m/s .

We have to calculate the radius ( r ) of of curve.

The centripetal acceleration ( a ) is given by

a = \frac{v^{2} }{r}

r = \frac{v^{2} }{a}  = \frac{40^{2} }{9.40}  = \frac{1600}{9.40} = 170.21 m

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Capacitor C1 is initially charged to V1 and capacitor C2 is initially charged to V2. The capacitors are then connected to each o
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Answer:

<em>20.08 Volts</em>

Explanation:

<u>Parallel Connection of Capacitors</u>

The voltage across any two elements connected in parallel is the same. If the elements are capacitors, then each voltage is

\displaystyle V_1=\frac{Q_1}{C_1}

\displaystyle V_2=\frac{Q_2}{C_2}

They are both the same after connecting them, thus

\displaystyle \frac{Q_2}{C_2}=\frac{Q_1}{C_1}

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