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Physics 18 Online
OpenStudy (anonymous):

B.find the magnitude and direction of the net force in each situation. 5. A 500 passenger boring 747 jet (with a mass of 1.6 x 105 kg) undergoes an acceleration of 1.2 m/s^2 [S] along a runway. 6. A cannon gives a 5.0 kg shell a forward acceleration of 5.0 x 103 m/s^2 before it leaves the muzzle. 7. A 28 g arrow is given an acceleration of 2.5 x 10^3 m/s^2. 8. the world's fastest coconut tree climber takes only 4.88 seconds to climb barefoot 8.99 meters up a coconut tree. calculate the climber's average velocity for this motion, assuming that the climb was vertically upward. @johnweldon1993 @ganeshie8 @ParthKohli @aaronq @kawaiicat123

OpenStudy (abhisar):

Hello @jansd345 !

OpenStudy (anonymous):

hi! @Abhisar

OpenStudy (abhisar):

In the first situation, two forces are acting on the jet. One is the thrust force and the other is gravitational force.

OpenStudy (abhisar):

Both forces are perpendicular to each other. Net force will be equal to the vector sum of each force. \(\sf \sqrt{(1.6\times 10^5\times 1.2)^2+(1.6\times 10^5\times 10)^2}\)

OpenStudy (abhisar):

I hope you know how to add vectors ?

OpenStudy (anonymous):

I'm still not sure with that yet

OpenStudy (anonymous):

@Abhisar

OpenStudy (abhisar):

Do you have to learn vectors in your syllabus ? Do you have answer key to these questions ?

OpenStudy (anonymous):

as I mentioned, in the syllabus, it's not very descriptive. and right now, I don't see on the lesson about vectors.. No, I don't have an answer key too... That's why I need help

OpenStudy (abhisar):

Ok, then i think you have to simply calculate the force in the given situations.

OpenStudy (abhisar):

Force = Acceleration \(\times\)Mass nad the direction of force is towards the direction of acceleration.

OpenStudy (abhisar):

*and

OpenStudy (abhisar):

Can you solve the questions now ?

OpenStudy (anonymous):

so the 1.6x105 kg do I times that? 1.6x105?

OpenStudy (anonymous):

I'm pretty sure I could, but I need a double check with a professional. physics isn't really my thing, and it gets really confusing..

OpenStudy (abhisar):

is it 1.6\(\times\)105 or 1.6\(\times 10^5\) ?

OpenStudy (anonymous):

no 105

OpenStudy (abhisar):

ok. for first question, Force = 1.6 X 105 X 1.2

OpenStudy (anonymous):

okay

OpenStudy (abhisar):

Calculate similarly for rest of the problems.

OpenStudy (abhisar):

Don't forget to keep everything in SI units

OpenStudy (anonymous):

wait, in these question I am finding the magnitude and direction?

OpenStudy (abhisar):

By the above method you are finding the magnitude. Direction will be same as that of acceleration. Like for first question direction of acceleration is given south, so the direction of force will also be south.

OpenStudy (abhisar):

Got it ?

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