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3 years ago

14

A car drives straight down toward the bottom of a valley and up the other side on a road whose bottom has a radius of curvature

of 115m. At the very bottom, the normal force on the driver is twice his weight. At what speed was the car traveling?

1 answer:

mina [271]3 years ago

6 0

Answer:33.58794 m/s

Explanation : fCenrtipetal force = mv2/r

m = Mass of car

r = Radius of curvature = 115 m

g = Acceleration due to gravity = 9.81 m/s²

The normal force is given by

N=Fc+mg

At the bottom N = 2N

2mg=fc+mg

⇒2mg=mv2

The car was traveling at the speed of 33.58794 m/s

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Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at

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Answer:

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Explanation:

given information:

$q_{1}$ = 3 nC = 3 x $10^{-9}$ C

$q_{2}$ = 2 nC = 2 x $10^{-9}$ C

r = 50 cm = 0.5 m

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F = k $\frac{q_{e}q}{r}$

where

k = constant (8.99 x $10^{9} Nm^{2} /C^{2}$ )

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since it is measured at the midpoint,

r = $\frac{0.5}{2}$

= 0.25 m

thus,

F = $F_{1}+ F_{2}$

= k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $\frac{kq_{e} }{r}$ ( $q_{1} +q_{2}$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ +2 x $10^{-9}$ )/0.25

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$r_{2}$ = 0.5 - 0.1 = 0.4 m

F = k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $kq_{e}$ ( $\frac{q_{1} }{r_{1} }$ + $\frac{q_{2} }{r_{2} }$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ /0.1+2 x $10^{-9}$ /0.4)

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