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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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viktelen [127]

Answer:

a)    F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} ),   b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

         F =k \frac{q_2q_2}{r^2}

         

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

          F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron       r₁ = x

right side -electron     r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

          F_net = k q Q  ( \frac{1}{r_1^2}+ \frac{1}{r_2^2})

          F _net = kqQ  ( \frac{1}{x^2} + \frac{1}{(d-x)^2} )

         

let's substitute the values

          F_net = 9 10⁹  1.6 10⁻¹⁹ 4.50 10⁻⁹ ( \frac{1}{x^2} + \frac{1}{(0.30-x)^2} )

          F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values ​​of the distance (x) and the net force are given

x (m)        F (N)

0.05        27.0 10-16

0.10          8.10 10-16

0.15          5.76 10-16

0.20         8.10 10-16

0.25        27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

4 0
3 years ago
Which type of surface would most likely be the best reflector of electromagnetic energy?
Len [333]
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5 0
2 years ago
What is net force?
kramer

Answer:

A. The sum of all the forces acting on an object.

3 0
3 years ago
Which amusement park ride utilizes periodic motion?
tatyana61 [14]
Any ride that oscillates back and forth or moves only in a complete circle utilizes periodic motion.
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2 years ago
A 1180kg car is moving at a speed of 85.5 km/h. Find the force needed to bring the car to rest in a distance of 300m
yarga [219]

Answer:

1110 N

Explanation:

First, find the acceleration.

Given:

Δx = 300 m

v₀ = 85.5 km/h = 23.75 m/s

v = 0 m/s

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v² = v₀² + 2aΔx

(0 m/s)² = (23.75 m/s)² + 2a (300 m)

a = -0.94 m/s²

Find the force:

F = ma

F = (1180 kg) (-0.94 m/s²)

F = -1110 N

The magnitude of the force is 1110 N.

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