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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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sod_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2=

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3 years ago
Can someone solve this problem and explain to me how you got it​
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Answer:

question5: F=74312.5N

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Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒F\alpha\frac{q1*q2}{r^{2}}

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