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Paul [167]
3 years ago
8

Mary and her younger brother Alex decide to ride the 17-foot-diameter carousel at the State Fair. Mary sits on one of the horses

in the outer section at a distance of 8 feet from the center. Alex decides to play it safe and chooses to sit in the inner section at a distance of 5 feet from the center.(a) What is Mary's angular speed ?M compared to that of Alex's angular speed ?A? Give your answer as a multiple of ?A.(b) What is Mary's tangential speed vM compared to that of Alex's tangential speed vA? Give your answer as a multiple of vA.
Physics
1 answer:
lorasvet [3.4K]3 years ago
5 0

Answer:

\omege_A=\omega_M

v_M=1.6v_A

Explanation:

A denotes Alex

M denotes Mary

r = Distance from center

Mary and Alex will have the equal displacements in equal interval of time as they are in uniform circular motion. So,

\omega_A=\omega_M

Tangential speed speed is given by

\dfrac{v_M}{v_A}=\dfrac{r_M\omega_M}{r_A\omega_A}\\\Rightarrow \dfrac{v_M}{v_A}=\dfrac{r_M}{r_A}\\\Rightarrow \dfrac{v_M}{v_A}=\dfrac{8}{5}\\\Rightarrow \dfrac{v_M}{v_A}=1.6\\\Rightarrow v_M=1.6v_A

The tangential speed of Mary is v_M=1.6v_A

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Answer:

a. the force between them quadruples

Explanation:

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3 years ago
What ocean depth would the volume of an aluminium sphere be reduced by 0.10%
yKpoI14uk [10]

Answer:

6400 m

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You need to use the bulk modulus, K:

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where ρ is density and P is pressure

Since ρ is changing by very little, we can say:

K ≈ ρ ΔP/Δρ

Therefore, solving for ΔP:

ΔP = K Δρ / ρ

We can calculate K from Young's modulus (E) and Poisson's ratio (ν):

K = E / (3 (1 - 2ν))

Substituting:

ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)

Before compression:

ρ = m / V

After compression:

ρ+Δρ = m / (V - 0.001 V)

ρ+Δρ = m / (0.999 V)

ρ+Δρ = ρ / 0.999

1 + (Δρ/ρ) = 1 / 0.999

Δρ/ρ = (1 / 0.999) - 1

Δρ/ρ = 0.001 / 0.999

Given:

E = 69 GPa = 69×10⁹ Pa

ν = 0.32

ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)

ΔP = 64.0×10⁶ Pa

If we assume seawater density is constant at 1027 kg/m³, then:

ρgh = P

(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa

h = 6350 m

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