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An auto weighing 2,500 pounds is on a street inclined at 10 with the horizontal. Find the force necessary to prevent the car from rolling down the hill.



Round your answer to the nearest whole number. Do not use a space or decimal in your answer or it will be marked wrong.

Answer :

Aliwohaish12
Make sure your calculator is set to DEGREES and not RADIANS when performing your sinθ calculation. sin(10) = -0.54402 sin(10°) = 0.17365 You could also try using trig tables if you're unsure how to do this. So: 2500sin(10°) = 434 lb

Answer:

13892 lb-ft/sec^2

Step-by-step explanation:

Given that An auto weighing 2,500 pounds is on a street inclined at 10 with the horizontal.

Here acceleration due to gravity is the pulling factor against which the car should be stopped from rolling down the hill.

Weight of the car =2500 pounds

Vertical component of gravity = 32cos (90-10)=32sin 10 = 32(0.1736)

Force required = mass x acceleration

=2500x0.1736x32

=13892 lb ft/sec^2

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