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viktelen [127]
3 years ago
13

A 39.5 kg child stands at the center of a 125 kg playground merry-go-round which rotates at 3.10 rad/s. If the child moves to th

e edge of the merry-go-round, what is the new angular velocity of the system? Model the merry-go-round as a solid disk
Physics
1 answer:
Neporo4naja [7]3 years ago
3 0
Using the law of conservation of angular momentum, we have 

<span>I1 w1 = I2 w2 </span>

<span>ie., m1r^2/2 x w1 = ( m1r^2/2 + m2r^2 ) w2 </span>

<span>ie., new angular velocity w2 = m1 w1 / ( m1+ 2m2) = 125 x 3.1 / ( 125 + 2 x39.5 ) </span>

<span>= 1.8995 = 1.9 rad /sec ( nearly )</span>
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A spring stretches 5 cm when a 300-N mass is suspended from it. Calculate the spring constant in N / m .
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Answer:

Spring constant in N / m = 6,000

Explanation:

Given:

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Spring constant in N / m

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Spring constant in N / m = Force/Distance

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Read 2 more answers
A car is running at the speed of 45km/hr see a child 25 meter ahead and suddenly apllies a brakes. If the retradation of the car
kenny6666 [7]

Answer:

The car stops in 7.78s and does not spare the child.

Explanation:

In order to know if the car stops before the distance to the child, you take into account the following equation:

x=x_o+v_ot-\frac{1}{2}at^2        (1)

vo: initial speed of the car = 45km/h

a: deceleration of the car = 2 m/s^2

t: time

xo: initial distance to the child = 25m

x: final distance to the child = 0m

It is necessary that the solution of the equation (1) for time t are real.

You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:

45\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=12.5\frac{m}{s}

0=25+12.5t-2t^2\\\\2t^2-12.5t-25=0\\\\t_{1,2}=\frac{-(-12.5)\pm \sqrt{(-12.5)^2-4(2)(-25)}}{2(2)}\\\\t_{1,2}=\frac{12.25\pm 18.87}{4}\\\\t_1=7.78s\\\\t_2=-1.65s

You take the first value t1 because it has physical meaning.

The solution for t is real, then, the car stops in 7.78s and does not spare the child.

4 0
3 years ago
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