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Nataly_w [17]
3 years ago
7

A turntable A is built into a stage for use in a theatrical production. It is observed during a rehearsal that a trunk B starts

to slide on the turntable 10 s after the turntable begins to rotate. Knowing that the trunk undergoes a constant tangential acceleration of 0.25 m/s2, determine the coefficient of static friction between the trunk and the turntable.
Physics
1 answer:
Ugo [173]3 years ago
3 0

Answer:

0.26

Explanation:

The radius of the table is 2.5 m.

Tangential velocity,

v=u+at\\u=0\\t=10s\\a=0.25m/s^2\\v=0+0.25\times 10 = 2.5 m/s

Net force on the trunk in the plane of the table:

F=m\sqrt{a^2+(\frac{v^2}{r})^2}\\F= m\sqrt{0.25^2+(\frac{2.5^2}{2.5})^2}=2.5m

The trunk slides because the frictional force equals the net force acting on the table:

\mu m g = 2.5 m\\\mu =\frac{2.5}{g} = 0.26

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Answer:

26621 km

Explanation:

We are given;

Mass: m = 5.98 x 10^(24) kg

Period; T = 43200 s

Formula for The velocity(v) of the satellite is:

v = 2πR/T

Where R is the radius

Formula for centripetal acceleration is;

a_c = v²/R

Thus; a_c = (2πR/T)²/R = 4π²R/T²

Formula for gravitational acceleration is:

a_g = Gm/R²

Where G is gravitational constant = 6.674 × 10^(-11) m³/kg.s²

Now the centripetal acceleration of the satellite is caused by its gravitational acceleration. Thus;

Centripetal acceleration = gravitational acceleration.

Thus;

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Making R the subject gives;

R = ∛(GmT²/4π²)

Plugging in the relevant values;

R = ∛((6.674 × 10^(-11) × 5.98 x 10^(24) × 43200²)/(4 × π²))

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Answer:

(a) Acceleration of female astronaut = 9.33*10^-12 m/s^2; Acceleration of male astronaut = 7.56*10^-12 m/s^2

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(c) No, their accelerations would not be constant.

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Mass of female astronaut M_{1} = 60.0 kg

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(a) The free-body diagram is shown in the attached figure.

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