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erica [24]
1 year ago
5

An elevator has a weight of 14,700 N and has an acceleration of –0.30 m/s2. The free-body diagram shows the forces acting on the

elevator.
A free body diagram with 2 force vectors. One vector is pointing up, labeled F Subscript t Baseline. The second vector is shorter pointing down, labeled F Subscript g Baseline.
To the nearest whole number, what is the force of tension, Ft, acting on the elevator?

Physics
2 answers:
attashe74 [19]1 year ago
6 0

Answer:

14,250 N

Explanation:

Weight = 14,700 N

Acceleration = -0.30m/s²

Acceleration = -0.30m/s2[tex]Weight=Fg\\\\Fg=mg\\\\Fg=14,700N\\m=?\\g=9.8m/s^{2} \\\\14,700N=m(9.8m/s^{2} ) \\\\14,700N/9.8ms^{2} =1500kg(mass)\\\\m=1500kg\\a=-0.30m/s^{2}\\\\f=ma\\f=(1500kg)(-0.30m/s^{2} )\\=-450N (net force)\\\\-450N=-14,700N+(Ft)\\14,250N=Ft

qwelly [4]1 year ago
4 0

14250. I just took it

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Given that,

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Using cosine law

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\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}

\alpha=\cos^{-1}(0.3846)

\alpha=67.38^{\circ}

We need to calculate the angle β

Using cosine law

50^2=130^2+120^2-2\times130\times120\cos\beta

\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}

\beta=\cos^{-1}(0.923)

\beta=22.63^{\circ}

We need to calculate the force on 130 cm side

Using formula of force

F_{130}=ILB\sin\theta

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We need to calculate the force on 120 cm side

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F_{120}=ILB\sin\beta

F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63

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The direction of force is out of page.

We need to calculate the force on 50 cm side

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F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38

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Answer:

Final speed of car = 12 m/s

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We have equation of motion v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time.

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        v = ?

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         t = 5 s

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b) Then maintains that velocity for 10 s

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         v = u + at = 20 + 0 x 10 = 20 m/s

c) Then decelerates at the rate of 2.0 m/s² for 4.0 s

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2 years ago
A force of 10 newtons toward the right is exerted on a
weeeeeb [17]

Answer:

Explanation:

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Answer:

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