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strojnjashka [21]
3 years ago
12

Skater begins to spend with arms held out at shoulder height. The skater wants to match the speed of the spin to the beat of the

music. How does the skater use knowledge of the conservation of momentum to do this?
Physics
1 answer:
Aleksandr [31]3 years ago
7 0

Answer:

the moment of inertia with the arms extended is Io and when the arms are lowered the moment

I₀/I > 1    ⇒   w > w₀

Explanation:

The angular momentum is conserved if the external torques in the system are zero, this is achieved because the friction with the ice is very small,

           L₀ = L_f

           I₀ w₀ = I w

          w =\frac{I_o}{I} w₀

where we see that the angular velocity changes according to the relation of the angular moments, if we approximate the body as a cylinder with two point charges, weight of the arms

          I₀ = I_cylinder + 2 m r²

where r is the distance from the center of mass of the arms to the axis of rotation, the moment of inertia of the cylinder does not change, therefore changing the distance of the arms changes the moment of inertia.

If we say that the moment of inertia with the arms extended is Io and when the arms are lowered the moment will be

        I <I₀

        I₀/I > 1    ⇒   w > w₀

therefore the angular velocity (rotations) must increase

in this way the skater can adjust his spin speed to the musician.

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tigry1 [53]
Around -3.5 ( might not be completely right ) Frank has a greater mass to he has a low velocity at first, Dion is around half of his weight so he doesn’t have to use as much force to get more than Frank
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3 years ago
Two cars A and B are moving with velocities 20 m/s and 15 m/s in the direction east and west respectively. If
Sphinxa [80]

Answer:

<u>Distance</u><u> </u><u>between</u><u> </u><u>them</u><u> </u><u>is</u><u> </u><u>4</u><u>,</u><u>2</u><u>0</u><u>0</u><u> </u><u>meters</u><u>.</u>

Explanation:

Consinder car A:

{ \bf{distance =  speed \times time }}

substitute:

distance = 20 \times (2 \times 60) \\  = 2400 \: m

Consider car B:

distance = 15 \times (2 \times 60) \\  = 1800 \: m

since these cars move in opposite directions, distance between them is their summation:

distance \: between = { \sum(distance \: of \: each \: car)} \\  = 2400 + 1800 \\  = 4200 \: m

3 0
3 years ago
swings a 5.5 kg cup of water in a vertical circle of radius 1.9 m. (a) What minimum speed must the cup have in this demo if the
Tanzania [10]

Answer:

4.32

Explanation:

The centripetal acceleration of any object is given as

A(cr) = v²/r, where

A(c) = the centripetal acceleration

v = the linear acceleration

r = the given radius, 1.9 m

Since we are not given directly the centripetal acceleration, we'd be using the value of acceleration due to gravity, 9.8. This means that

9.8 = v²/1.9

v² = 1.9 * 9.8

v² = 18.62

v = √18.62

v = 4.32 m/s

This means that, the minimum speed the cup must have so as not to get wet or any spill is 4.32 m/s

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A very weak pressure wave, i.e., a sound wave, across which the pressure rise is 30 Pa moves through air which has a temperature
Fofino [41]

Answer:

Density change, Δρ = 2.4 × 10⁻⁴ kg/m³

Temperature Change, ΔT = 0.0258 K

Velocity Change, Δc = 0.0148 m/s

Explanation:

For sound waves moving through the air,

Pressure and Temperature varies thus

(P₀/P) = (T₀/T)^(k/(k-1))

Where P₀ = initial pressure of air = 101KPa = 101000 Pa

P = final pressure of air due to the change brought about by the moving sound wave = 101000+30 = 101030 Pa

T₀ = initial temperature of air = 30°C = 303.15 K

T = final temperature of air = ?

k = ratio of specific heats = Cp/Cv = 1.4

(101000/101030) = (303.15/T)^(1.4/(1.4-1))

0.9990703 =(303.15/T)^(3.5)

Solving This,

T = 303.1758 K

ΔT = T - T₀ = 303.1758 - 303.15 = 0.0258 K

Density can be calculate in two ways,

First method

Δρ = ρ - ρ₀

P₀ = ρ₀RT₀

ρ₀ = P₀/RT₀

R = gas constant for air = 287 J/kg.k

where all of these are values for air before the wave propagates

P₀ = 101000 Pa, R = 287 J/kg.K, T₀ = 303.15K

ρ₀ = 101000/(287 × 303.15) = 1.1608655 kg/m³

ρ = P/RT

P = 101030 Pa, T = 303.1758K

ρ = 101030/(287×303.1758) = 1.1611115 kg/m³

Δρ = ρ - ρ₀ = 1.1611115 - 1.1608655 = 0.00024 kg/m³ = 2.4 × 10⁻⁴ kg/m³

Second method

(ρ₀/ρ) = (T₀/T)^(1/(k-1))

Where ρ₀ is initially calculated from ρ₀ = P₀/RT₀, then ρ is then computed and the diff taken.

Velocity Change

c₀ = √(kRT₀) = √(1.4 × 287 × 303.15) = 349.00669 m/s

c = √(kRT) = √(1.4 × 287 × 303.1758) = 349.0215415 m/s

Δc = c₀ - c = 349.0215415 - 349.00669 = 0.0148 m/s

QED!

5 0
3 years ago
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