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3 years ago

7

A trapeze artist, with swing, weighs 800 N; he is momentarily heldto one side by his partner using a horizontal force so that th

eswing ropes make an angle of 30o with thevertical. In such a condition of static equilibrium, what isthe horizontal force being applied by the partner?

1 answer:

Tamiku [17]3 years ago

6 0

Answer:

461.88 N

Explanation:

$F_{g}$ = Weight of the swing = 800 N

$T$ = Tension force in the rope

$F$ = Horizontal force being applied by the partner

Using equilibrium of force in vertical direction using the force diagram, we get

$T Cos30 = F_{g}\\T Cos30 = 800\\T = \frac{800}{Cos30} \\\\T = 923.76 N$

Using equilibrium of force in horizontal direction using the force diagram, we get

$F = T Sin30\\F = (923.76) (0.5)\\F = 461.88 N$

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8. A 60 kg runner has 1500 J of kinetic energy. How fast is he moving?

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Answer: 7.07 m/s

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Speed of runner = ?

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A series AC circuit contains a resistor, an inductor of 220 mH, a capacitor of 5.50 μF, and a generator with ΔVmax = 240 V opera

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Answer:

$Xl=wL=69.11503383 ohm$

$Xc=\frac{-1}{wC}= -578.7452476 ohm$

$Z=\frac{240}{140*10^-3}=1.714285714 kilo-ohm$

$R=2.223915928 kilo-ohm$

$\alpha = -12.90698798 centigrades$

Explanation:

$L=220*10^-3$

$C=5.50*10^-6$

$Vmax=240V$

$Imax=140*10^-3$

$w=2\pi f=2\pi *50=314.1592654$

The capacitive reactance is given by:

$Xc=\frac{-1}{wC}=-578.7452476 ohm$

Now, The inductive reactance is given by:

$Xl=wL=69.11503383 ohm$

By the ohm´s law, the electrical impedance is:

$V=IZ$

So

$Z=\frac{V}{I}$

$Z=\frac{240}{140*10^-3}=1.714285714 kilo-ohm$

The total impedance is:

$Z=R+jX\\\\$ (*)

Where X is the total reactance given by:

$X=Xl+Xc=69.11503383-578.7452476=-509.6302138$

Let´s calculate the real part of Z using (*):

$R=1714.285714+509.6302138$

$R=2.223915928 kilo-ohm$

Finally the angle between the current and the voltage is equal to the impedance angle:

$\alpha=arctan(\frac{-509.6302138}{2223.915928})$

$\alpha =-12.90698798 centigrades$

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