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tekilochka [14]
4 years ago
10

A tetherball is attached to a pole with a 2.0-m rope. It is circling at 0.20 rev/s. As the rope wraps around the pole it shorten

s. How long is the rope when the ball is moving at 5.0 m/s?
Physics
1 answer:
denpristay [2]4 years ago
5 0

Answer:

The length of the rope  is 1.005 m.

Explanation:

Given that,

Length of rope = 2.0 m

Angular speed = 0.20 rev/s

Linear speed = 5.0 m/s

Since there are no external torques about the rotation  axis

We need to calculate the length of the rope

Using conservation of angular momentum

I_{1}\omrga_{1}=I_{2}\omega_{2}

mr_{1}^2\times\omega_{1}=mr_{2}^2\times\omega_{2}

r_{1}^2\times\omega_{1}=r_{2}^2\times\dfrac{v_{2}}{r_{2}}

r_{2}=\dfrac{r_{1}^2\times\omega_{1}}{v_{2}}

Put the value into the formula

r_{2}=\dfrac{2.0^2\times0.20\times2\pi}{5.0}

r_{2}=1.005\ m

Hence, The length of the rope  is 1.005 m.

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Answer:

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b) T = 608.22 N

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d) T = 533.82 N

Explanation:

Given that the mass of gymnast is m = 62.0 kg

Acceleration due to gravity is g = 9.81 m/s²

Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.

So;

To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;

T = mg

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs the rope at a constant rate tension in the string is

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs up the rope with an upward acceleration of magnitude

a = 1.2 m/s²

the tension in the string is  T - mg = ma (Since acceleration a is upwards)

T = ma + mg

= m (a + g )

= (62.0 kg)(9.81 m/s² + 1.2  m/s²)

= (62.0 kg) (11.01 m/s²)

= 682.62 N

When the gymnast climbs up the rope with an downward acceleration of magnitude

a = 1.2 m/s² the tension in the string is  mg - T = ma (Since acceleration a is downwards)

T = mg - ma

= m (g - a )

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= (62.0 kg)(8.61 m/s²)

= 533.82 N

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Explanation:

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A car moves with constant acceieration of -0.s m/s? on a straight portion of the road. Att- 0s the car has a velocity of 69 mph,
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Answer:

b) d = 0.71 Km

Explanation:

Car kinematics

Car 1 moves with uniformly accelerated movement

v_f^2=v_o^2+2a*d Formula (1)

d: displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Equivalences:

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1 hour = 3600s

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Known data

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a = -0.5 m/s²

Distance calculation

We replace data in the Formula (1)

15.4^2 = 30.8^2+2(-0.5)*d

2(0.5)*d = 30.8^2 - 15.4^2

d =\frac{ 30.8^2 - 15.4^2}{2(0.5) }= 717.6m

d = 717.6 m\frac{1km}{1000m} = 0.7176Km

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