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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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Answer:

0.38°

Explanation:

\theta = Angle

m = Number

d = Distance

n = Refractive index of liquid = 1.25

a denotes air

l denotes liquid

In the case of double split interferance we have the relation

m\lambda=dsin\theta

For air

m\lambda_a=dsin\theta_a

For liquid

m\lambda_l=dsin\theta_l

Dividing the two equations

\frac{m\lambda_a}{m\lambda_l}=\frac{dsin\theta_a}{dsin\theta_l}\\\Rightarrow \frac{\lambda_a}{\lambda_l}=\frac{sin\theta_a}{sin\theta_l}

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The angular separation is 0.38°

5 0
3 years ago
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One mole of iron (6 x 10^23 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-
choli [55]

Answer:

Explanation:

Given that:

length l = 2.3 m

a = 0.12 cm = 0.12  \times 10^{-2} \ m

x = 1.17 \ cm = 1.17 \times 10^{-2}\ m

m = 149 kg

\delta = 7.87 \ g/cm^3

da = 2.28 \times 10^{-10}\ m

F_{net} = F-mg\\ \\0 = F - mg \\ \\  F = mg \\ \\ k_sx = mg \\ \\

∴

k_s = \dfrac{149(9.8)}{1.17 \times 10^{-2}} \\ \\  k_s = 124803.42  \ N /m

N_{chain} = \dfrac{A_{wire}}{A_{atom}} = \dfrac{A_w}{da^2}

N_{chain} = \dfrac{(a)^2}{(da)^2} = (\dfrac{a}{da})^2

N_{chain} =  (\dfrac{0.12 \times 10^{-2} }{2.28 \times 10^{-10}})^2

N_{chain} = 2.77 \times 10^{13}

N_{bond} = \dfrac{L}{da} \\ \\  = \dfrac{2.3}{2.28 \times 10^{-10}} \\ \\ N_{bond} = 1.009 \times 10^{10}

\text{Finally; the stiffness of a single interatomic spring is:}

k_{si} =\dfrac{N_{bond}}{N_{chain}}\times k_s

k_{si} =\dfrac{(1.009 \times 10^{10})}{2.77*10^{13}}}\times (124803.42)

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2 years ago
A 20 kg crate initially at rest on a horizontal floor requires a 80 N horizontal force to set it in motion. Find the coefficient
e-lub [12.9K]

Answer:

<em>The coefficient of static friction between the crate and the floor is 0.41</em>

Explanation:

<u>Friction Force</u>

When an object is moving and encounters friction in the air or rough surfaces, it loses acceleration and velocity because the friction force opposes motion.

The friction force when an object is moving on a horizontal surface is calculated by:

Fr=\mu N          [1]

Where \mu is the coefficient of static or kinetics friction and N is the normal force.

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N = W = m.g

The crate of m=20 Kg has a weight of:

W = 20*9.8

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The normal force is also N=196 N

We can find the coefficient of static friction by solving [1] for \mu:

\displaystyle \mu=\frac{Fr}{N}

The friction force is equal to the minimum force required to start moving the object on the floor, thus Fr=80 N and:

\displaystyle \mu=\frac{80}{196}

\mu=0.41

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