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Dvinal [7]
3 years ago
12

A diver can reduce her moment of inertia by a factor of about 3.5 when changing from the straight position to the tuck position.

If she makes two rotations in 1.5 seconds when in the tuck position, what is her angular speed (rad/sec) when in the straight position?
Physics
1 answer:
katovenus [111]3 years ago
4 0

Answer:

29.3 rad/s

Explanation:

Moment of inertia in straight position, I1 = I

He takes 2 rotations in 1.5 second

So, Time period, T = 1.5 / 2 = 0.75 second

w1 = 2 x 3.14 / T = 2 x 3.14 / 0.75 = 8.37 rad/s

Moment of inertia in tucked position, I2 = I / 3.5

Let the new angular speed is w2.

By the use of conservation of angular momentum, if no external torque is applied, then angular momentum is constant.

L1 = L2

I1 w1 = I2 w2

I x 8.37 = I / 3.5 x w2

w2 = 29.3 rad/s

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If the period of the pendulum is tripled, what is the length of the string increased by?
Nitella [24]
The period of the pendulum doesn't determine the length of the string. 
It's the other way around.

The period of the pendulum is proportional to the square root of its length.
So if you want to triple the period, you have to make the string nine times
as long as it is now.
7 0
3 years ago
An interference pattern is produced by light with a wavelength 580 nm from a distant source incident on two identical parallel s
irakobra [83]

Answer:

Explanation:

1 )

Here

wave length used that is λ = 580 nm

=580 x 10⁻⁹

distance between slit d = .46 mm

= .46 x 10⁻³

Angular position of first order interference maxima

= λ / d radian

= 580 x 10⁻⁹ / .46 x 10⁻³

= 0.126 x 10⁻² radian

2 )

Angular position of second order interference maxima

2 x  0.126 x 10⁻² radian

= 0.252 x 10⁻² radian

3 )

For intensity distribution the formula is

I = I₀ cos²δ/2 ( δ is phase difference of two lights.

For angular position of θ1

δ = .126 x 10⁻² radian

I = I₀ cos².126x 10⁻²/2

= I₀ X .998

For angular position of θ2

I = I₀ cos².126x2x 10⁻²/2

=  I₀ cos².126x 10⁻²

8 0
3 years ago
From the top of a cliff, a person uses a slingshot to fire a pebble straight downward, which is the negative direction. The init
Talja [164]

Answer:

\vec{a} = -(9.8~{\rm m/s^2})\^y

Explanation:

Regardless of the initial velocity of the pebble, the acceleration of the pebble is equal to the gravitational acceleration which is equal to 9.8 m/s2 towards downwards direction.

This can be shown by Newton's Second Law. According to the law, the net force applied on an object is equal to mass times acceleration of that object.

During the downward motion, the only force acting on the pebble is the gravitational force, hence its acceleration is equal to gravitational acceleration.

8 0
3 years ago
A uniform meter rule of weight 0.9N is suspended horizontally by two vertical loops of thread A and B placed 20cm and 30cm from
Rzqust [24]

The  distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

<h3>Distance from the center of the meter rule</h3>

The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;

-----------------------------------------------------------------

  20 A  (30 - x)↓      x         ↓      20 cm  B 30 cm

                       2N              0.9N

Let the center of the meter rule = 50 cm

take moment about the center;

2(30 - x) + 0.9(x)(30 - x) = 0.9(20)

(30 - x)(2 + 0.9x) = 18

60 + 27x - 2x - 0.9x² = 18

60 + 25x - 0.9x² = 18

0.9x² - 25x - 42 = 0

x = 29.3 cm

Thus, the  distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

Learn more about brainly.com/question/874205 here:

#SPJ1

7 0
2 years ago
Read 2 more answers
A circus tightrope walker weighing 800 N is standing in the middle of a 15 meter long cable stretched between two posts. The cab
Lorico [155]

Answer:

T = 10010 N

Explanation:

To solve this problem we must use the translational equilibrium relation, let's set a reference frame

X axis

       Fₓ-Fₓ = 0

       Fₓ = Fₓ

whereby the horizontal components of the tension in the cable cancel

Y Axis  

        F_{y} + F_{y} - W =0

        2F_{y} = W

let's use trigonometry to find the angles

        tan θ = y / x

        θ = tan⁻¹ (0.30 / 0.50 L)

        θ = tan⁻¹ (0.30 / 0.50 15)

        θ = 2.29º

the components of stress are

         F_{y} = T sin θ

we substitute

       2 T sin θ = W

       T = W / 2sin θ

        T = \frac{ 800}{ 2sin 2.29}

        T = 10010 N

4 0
3 years ago
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