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11111nata11111 [884]

3 years ago

8

In the Skycoaster amusement park ride, riders are suspended from a tower by a long cable. A second cable then lifts them until t

hey reach the starting position indicated in (Figure 1). The lifting cable is then released, and the riders swing down the circular arc shown.

2 answers:

Alborosie3 years ago

4 0

Answer:

T1 cos 15 = T2 cos 26

T1 = T2 cos 26 / cos 15

T1 = 0.930 T2 ------------(1)

T1 sin 15 + T2 sin 26 = m g

substitute eqn(1)

(0.930 T2 ) sin 15 + T2 sin 26 = 230 * 9.8

0.241 T2 + 0.438 T2 = 2254

Tension in right cable T2 = 3319 N

substitute in eqn (1)

T1 = 0.930 * 3319

Tension in left cable T1 = 3087

inna [77]3 years ago

3 0

Answer:

T1 cos 15 = T2 cos 26

T1 = T2 cos 26 / cos 15

T1 = 0.930 T2 ------------(1)

T1 sin 15 + T2 sin 26 = m g

substitute eqn(1)

(0.930 T2 ) sin 15 + T2 sin 26 = 230 * 9.8

0.241 T2 + 0.438 T2 = 2254

Tension in right cable T2 = 3319 N

substitute in eqn (1)

T1 = 0.930 * 3319

Tension in left cable T1 = 3087 N

Explanation:

If the four riders have a total mass of 230 kg, what is the tension in the left cable just before release?

Express your answer with the appropriate units.

15° 26°

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8 0

3 years ago

An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a

Zolol [24]

Answer:

a) $\omega = 0.421\,\frac{rev}{s}$ , b) $\Delta \theta = 0.066\,rev$ , c) $v = 0.966\,\frac{mm}{s}$ , d) $a = 3.293\,\frac{mm}{s^{2}}$

Explanation:

a) The angular velocity of the turntable after $0.200\,s$ .

$\omega = \omega_{o} + \alpha\cdot \Delta t$

$\omega = 0.240\,\frac{rev}{s} + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)$

$\omega = 0.421\,\frac{rev}{s}$

b) The change in angular position is:

$\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot \alpha \cdot t^{2}$

$\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}$

$\Delta \theta = 0.066\,rev$

c) The tangential speed of a point on the rim of the turn-table:

$v = r\cdot \omega$

$v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$v = 9.655\times 10^{-4}\,\frac{m}{s}$

$v = 0.966\,\frac{mm}{s}$

d) The tangential and normal components of the acceleration of the turn-table:

$a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )$

$a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{t} = 2.078\,\frac{mm}{s}$

$a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}$

$a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{n} = 2.554\,\frac{mm}{s^{2}}$

The magnitude of the resultant acceleration is:

$a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}$

$a = 3.293\,\frac{mm}{s^{2}}$

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3 years ago

A pilot wishes to flt his planne to an airport north of his current location

Ann [662]

Answer:

Time = 2758.62 seconds

Explanation:

Given the following data;

Speed = 290 m/s

Distance = 800 km to meters = 800 * 1000 = 800000

To find the time;

Time = distance/speed

Time = 800000/290

Time = 2758.62 seconds

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