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Sergio039 [100]
3 years ago
10

A skater extends her arms horizontally, holding a 5-kg mass in each hand. She is rotating about a vertical axis with an angular

velocity of one revolution per second. If she drops her hands to her sides, what will the final angular velocity (in rev/s) be if her moment of inertia remains approximately constant at 5 kg⋅m2, and the distance of the masses from the axis changes from 1 m to 0.1 m?

Physics
2 answers:
Crazy boy [7]3 years ago
7 0

The answer is attached

sp2606 [1]3 years ago
3 0

Answer:

1.98 rev/s

Explanation:

m  = mass attached to each hand = 5 kg

r_{i} = initial distance of masses in each hand = 1 m

r_{f} = final distance of masses in each hand = 0.1 m

I = moment of inertia of body = 5 kgm²

I_{i} = initial total moment of inertia = I + 2 mr_{i}^{2}

w_{i} = initial angular velocity = 1 rev/s

I_{f} = final total moment of inertia = I + 2 mr_{f}^{2}

w_{f} = final angular velocity = ?

Using conservation of angular momentum

I_{i} w_{i} = I_{f} w_{f}

(I + 2 mr_{i}^{2}) w_{i} = (I + 2 mr_{f}^{2}) w_{f}

(5 + 2 (5) (1)^{2}) (1) = (5 + 2 (5) (0.1)^{2}) w_{f}

w_{f} = 2.94 rev/s

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Answer:

a.Attractive

2.31531\times 10^{-16}\ N

Explanation:

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Here, there is a proton and electron which are different particles hence, they will attract each other.

q_1=q_2 = Charge of electron and proton = 1.6\times 10^{-19}\ C

r = Distance between them = 997 nm

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Force is given by

F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(997\times 10^{-9})^2}\\\Rightarrow F=2.31531\times 10^{-16}\ N

The force of attraction between the particles will be 2.31531\times 10^{-16}\ N

8 0
3 years ago
If the tension in the rope is 160 n, how much work does the rope do on the skier during a forward displacement of 270 m?
Lunna [17]

If the tension in the rope is 160 n, - 43200 J work doen by the rope on the skier during a forward displacement of 270 m.

Given,

Tension force in the rope is (T) = 160 N

Displacement of the skier (S) = 270 m

The displacement takes place in forward direction while the direction of the tension in the rope is opposite to it.

Therefore, work done by the rope on  the skier is,

   W=T.S

⇒W=270*160*cos\pi \\W=-43200 J

Hence work done by the rope is - 43200 J.

Learn more about force problems on

brainly.com/question/26850893

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8 0
2 years ago
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Two particles with charges are initially very far apart (effectively an infinite distance apart). They are then fixed at positio
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Answer:

potential energy increases.

Explanation:

The potential energy between the two charged particles is given by

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If they are very far apart then r tends to infinity and the potential energy is zero.

If they come closer then the potential energy between the two charged particles increases.

Thus, the potential energy increases.

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3 years ago
On Earth, a brick has a mass of 10 kg and a weight of 5 lbs. What predictions could we make about the mass and weight of the bri
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Answer:

Mass remains constant but weight reduces

Explanation:

Mass is the amount of matter in an object so whether on moon or any other planet, it does not change despite the changes in acceleration.

Weight is a product of mass and acceleration due to gravity, expressed as W=mg where m is the mass, W is weight and g is acceleration. From the above formula, it is evident that when you decrease g, then W also decreases while m is constant. Similarly, when m is constant and g is increased then W also increases.

Therefore, for this case, since g decreases, the weight decreases but mass remains constant.

8 0
3 years ago
A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o
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Answer:

E = 31.329 N/C.

Explanation:

The differential electric field dE at the center of curvature of the arc is

dE = k\dfrac{dQ}{r^2}cos(\theta ) <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where r is the radius of curvature.

Now

dQ = \lambda rd\theta,

where \lambda is the charge per unit length, and it has the value

\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.

Thus, the electric field at the center of the curvature of the arc is:

E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta  }{r^2} cos(\theta)

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.

Now, we find \theta_1 and \theta_2. To do this we ask ourselves what fraction is the arc length  3.0 of the circumference of the circle:

fraction = \dfrac{3.0m}{2\pi (2.3m)}  = 0.2076

and this is  

0.2076*2\pi =1.304 radians.

Therefore,

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.

evaluating the integral, and putting in the numerical values  we get:

E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\

\boxed{ E = 31.329N/C.}

4 0
3 years ago
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