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Readme [11.4K]
3 years ago
5

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.250 rev/s . The magnitude

of the angular acceleration is 0.896 rev/s2 . Both the the angular velocity and angular accleration are directed counterclockwise. The electric ceiling fan blades form a circle of diameter 0.760 m. Compute the fan's angular velocity magnitude after time 0.207 s has passed.
Physics
1 answer:
Tems11 [23]3 years ago
5 0

Answer:

ω = 0.436 rev/sec = 2.736 rad/sec

Explanation:

By definition, the angular acceleration is the rate of change of angular velocity, which is described by the following expression:

\alpha = \frac{wf-wo}{t-to}

Solving for ωf :

ωf = ω₀ + α*t (assuming t₀ = 0)

Replacing by givens in the problem, ω₀ = 0.250 rev/s, t =0.207 sec, and α = 0.896 rev/s², we have:

ωf = 0.250 rev/s +( (0.896 rev/s²)*(0.207 s)) = 0.436 rev/sec

Expressed in rad/sec:

ωf = 0.436 rev/sec*2*\pi =2.736 rad/sec

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You rub a clear plastic pen with wool, and observe that a strip of invisible tape is attacted to the pen. Assuming that the pen
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option B

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Due to rubbing, the pen gets negatively charged.

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So, when the pen is negatively charged the tape might be positively charged or the tape might be uncharged.

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3 years ago
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
3 years ago
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