215.4K/s20۱۲:۵۲ صQuiz 1 Job.pdf >01:If a car is traveling at (80 km/h),determine its speed in miles per hour andfoots per second.80 km/h = {(80) (1000) / (0.3048)} / (5280) = 49.71 mi/h49.71 mi/h = (49.71) (5280) / (3600) = 72.91 ft/sO 2:Determine the magnitude of the force (F) sothat the resultant (FR) of the three forces is as30400 Nsmall as possible. What is the minimummagnitude of (FR)?500 NFR ΣFFRx = E Fx = F1x + F2x + F3x= 500 – F sin 30°– 400 sin 30° = (300 – 0.5F) NFRy = E Fy = Fıy + F2y + F3y= 400 cos 30°- F cos 30° +0= (346.4 – 0.866 F) NFR? = ( FRx)? + ( FRy )? = ( 300 – 0.5F )? + ( 346.4 – 0.866 F )?To find the minimum, we need to take the derivative of the magnitude and set it equal to zero.Take the derivative with respect to (F) so :dFR= 2 (300 – 0.5F) (- 0.5) + 2 (346.4 – 0.866 F) (- 0.866) = 02FR2 (- 150 + 0.25 F + 0.75 F – 300) = 0F= 450 NThe magnitude of the resultant force is:|FR| =( 300 – 0.5(450) ) ² + (346.4 – 0.866 (450) )²= (75 ) 2 + (-43.3 )² = 86.6 N

1 km = 0.62miles 1km = 3280.62foots 1hour = 3600sec. Just put these values to get the results

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01: If a car is traveling at (80 km/h), determine its speed in miles per hour and foots per second.

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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215.4
K/s
20
۱۲:۵۲ ص
Quiz 1 Job.pdf >
01:
If a car is traveling at (80 km/h),
determine its speed in miles per hour and
foots per second.
80 km/h = {(80) (1000) / (0.3048)} / (5280) = 49.71 mi/h
49.71 mi/h = (49.71) (5280) / (3600) = 72.91 ft/s
O 2:
Determine the magnitude of the force (F) so
that the resultant (FR) of the three forces is as
30
400 N
small as possible. What is the minimum
magnitude of (FR)?
500 N
FR ΣF
FRx = E Fx = F1x + F2x + F3x
= 500 – F sin 30°– 400 sin 30° = (300 – 0.5F) N
FRy = E Fy = Fıy + F2y + F3y
= 400 cos 30°- F cos 30° +0= (346.4 – 0.866 F) N
FR? = ( FRx)? + ( FRy )? = ( 300 – 0.5F )? + ( 346.4 – 0.866 F )?
To find the minimum, we need to take the derivative of the magnitude and set it equal to zero.
Take the derivative with respect to (F) so :
dFR
= 2 (300 – 0.5F) (- 0.5) + 2 (346.4 – 0.866 F) (- 0.866) = 0
2FR
2 (- 150 + 0.25 F + 0.75 F – 300) = 0
F= 450 N
The magnitude of the resultant force is:
|FR| =
( 300 – 0.5(450) ) ² + (346.4 – 0.866 (450) )²
= (75 ) 2 + (-43.3 )² = 86.6 N

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