anonymous
  • anonymous
A 1,800 kg car is parked on a road that has an elevation angle of 7°. Suppose the coefficient of static friction of the kinds of rubber and asphalt involved is 0.65. Which is approximately the force of static friction between the tires and the road? Question options: 2,200 N 9,300 N 11,350 N 18,000 N I got A, but I was told I was wrong.
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Fandude727
  • Fandude727
B?
anonymous
  • anonymous
How did you get that @Fandude727 ?
Fandude727
  • Fandude727
wait

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Fandude727
  • Fandude727
F = 1800kg x 9.80N/kg x sin7° F = 2150.0N I think A is right
anonymous
  • anonymous
Haha @Fandude727 That's exactly what I thought, but I was told I was wrong!
Fandude727
  • Fandude727
but it doesn't match any of the options
anonymous
  • anonymous
Yep, and 2200 isn't correct either. :(
Fandude727
  • Fandude727
Hmm,
Fandude727
  • Fandude727
Well, I'm completely stumped
anonymous
  • anonymous
@Fandude727 well thank you for trying :) I'm very stumped too haha!
Fandude727
  • Fandude727
You're welcome.
Fandude727
  • Fandude727
can you retake that question?
anonymous
  • anonymous
Hmm? I believe so, and I really just want to know where I went wrong.
Fandude727
  • Fandude727
So do I.
Fandude727
  • Fandude727
@Artkid101 My brother says it's D.
anonymous
  • anonymous
@Fandude727 How did he get that? Do you know?
Fandude727
  • Fandude727
No, he sat down at the computer for like 10 minutes and said it was D.
anonymous
  • anonymous
@Fandude727 Thank you so much, and thank you to your brother! :)
Fandude727
  • Fandude727
You're welcome, I'm glad you got it to work!
anonymous
  • anonymous
It was c :)
Fandude727
  • Fandude727
Awesome, glad to help you! :D

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