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Alexandra [31]
3 years ago
11

An electric fan is set on its HIGH setting. After the Low push button is depressed, the angular speed of the fan blades decrease

s to a value of 8.00x10^2 rev/min in 1.75sec. The deceleration is 42.0 rad/s^2. Determine the initial angular speed of the blades in rev/min
Physics
1 answer:
brilliants [131]3 years ago
8 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

<span>a=Δω/Δt
</span><span>a=2π*Δf/Δt
</span><span>a=2π*(f2-f1)/Δt
</span>
<span>f1=f2-a*Δt/2π
</span><span>f2=800/60 rev/sec
</span><span>a=-42 rad/sec^2
</span><span>Δt=1.75sec
</span><span>so f1=25 rev/sec
 f1=1500 rev/min</span>
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After being struck by a bowling ball, a 1.8 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.8 kg
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b) v₂ = 5 m/s

Explanation:

a)

We will use the law of conservation of momentum here:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0.8 m/s

v₂ = speed of second after before collsion = ?

Therefore,

(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s

<u>v₂ = 4.2 m/s</u>

<u></u>

b)

We will use the law of conservation of momentum here:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)

<u>v₂ = 5 m/s</u>

5 0
3 years ago
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