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Pavel [41]
3 years ago
13

A car travels along a straight line at a constant speed of 56.0 mi/h for a distance d and then another distance d in the same di

rection at another constant speed. The average velocity for the entire trip is 33.5 mi/h. (a) What is the constant speed with which the car moved during the second distance d?
I started out with Vxavg=(Vxi+Vxf)/2 and came out to 11 and it was wrong. If anyone has a good website I can use to learn physics it would be much appreciated.
Physics
1 answer:
ioda3 years ago
6 0
2* v1 * v2 / (v2+v1) = V avg2*56* v2 / (v2+56) = 33.5
v2 = 23.94 m/s
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Answer:

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Explanation:

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Replacing the given values, where k is the Coulomb constant:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N

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Answer:

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