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Scorpion4ik [409]
3 years ago
6

An engine flywheel initially rotates counterclockwise at 6.55 rotations/s. Then, during 20.9 s, its rotation rate changes to 2.1

9 rotations/s clockwise. Find the flywheel's average angular acceleration (including its sign) in radians per second squared. Define counterclockwise rotation as positive.
Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
7 0

Answer:

The average angular acceleration is -2.628 rad/s²

Explanation:

Counterclockwise = positive

Clockwise = -negative

Given;

initial rotation of the flywheel, θ₁ = 6.55 rotation/s

final rotation of the flywheel, θ₂ = - 2.19 rotation/s

The average angular acceleration is given by;

\alpha = \frac{\delta \theta}{\delta t}\\\\ \alpha =\frac{\theta _2 - \theta_ 1}{t}\\\\ \alpha =\frac{-2.19 -6.55}{20.9} \\\\ \alpha =\frac{-8.74}{20.9}\\\\ \alpha = -0.4182 \ rotation / s^2\\\\ \alpha = \frac{-0.4182 \ rotation}{s^2}*\frac{2\pi \ radian}{rotation}\\\\ \alpha = -2.628 \ rad/s^2

Therefore, the average angular acceleration is -2.628 rad/s²

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A watering can is most likely measured in which metric unit
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3 years ago
How fast was the storm that hit Galveston and how men died?​
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3 0
2 years ago
A light, flexible cable is wrapped around a solid cylinder with mass 3.3 kg and a radius of 0.8 meters. The cylinder rotates on
kari74 [83]

Answer:

9.16rad/s^2

Explanation:

We are given that

Mass,m_1=3.3 kg

Radius,r=0.8 m

m_2=4.9 kg

Height,h=2.9 m

We have to find the angular acceleration of the cylinder.

According to question

4.9g-T=4.9a

Tr=I\alpha

Where

\alpha=\frac{a}{r}

Tr=\frac{1}{2}m_1ra

T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a

Substitute the value

4.9g-\frac{1}{2}(3.3a)=4.9a

4.9\times 9.8=4.9a+\frac{3.3a}{2}

Where g=9.8 m/s^2

48.02=a(4.9+1.65)=6.55a

a=\frac{48.02}{6.55}=7.33m/s^2

Angular acceleration,\alpha=\frac{a}{r}=\frac{7.33}{0.8}=9.16rad/s^2

7 0
3 years ago
A fireman standing on a 15 m high ladder operates a water hose with a round nozzle of diameter 2.02 inch. The lower end of the h
Nataly_w [17]

Answer:

v₁ = 1,606 10⁴ m / s

Explanation:

For this exercise we must use Bernoulli's equation, let's use index 1 for the nozzle on the stairs and index 2 the pump on the street

              P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² +ρ g y₂

             

The pressure when the water comes out is the atmospheric pressure

           P₁ = P_atm

The difference in height between the street and the nozzle on the stairs is

           y₂-y₁ = 15 m

Now let's use the continuity equation

             v₁ A₁ = v₂ A₁

The area of ​​a circle is

             A = π r² = π (d/2)²

            v₁ π d₁²/ 4 = v₂ π d₂²/ 4

            v₂ = v₁ d₁² / d₂²

Let's replace

           P₂-P_atm + ½ ρ [ (v₁ d₁² / d₂²)²- v₁² ] + ρ g (y2-y1) = 0

           P₂- P_Atm + ρ g (y₂-y₁) = ½ ρ v₁² [1- (d₁/d₂)⁴]

           v₁² [1- (d₁/d₂)⁴] = (P₂-P_atm) ρ / 2 + g (y₂-y₁) / 2

Let's reduce the magnitudes to the SI system

           d₂ = 3.37 in (2.54 10⁻² m / 1 in) = 8.56 10⁻² m

           d₁ = 2.02 in = 5.13 10⁻² m

Let's calculate

            v₁² [1- (5.13 / 8.56) 4] = 449.538 10³ 10³/2  -9.8 15/2

            v₁² [0.8710] = 2.2477 10⁸ - 73.5 = 2.2477 10⁸

            v₁ = √ 2.2477 10⁸ /0.8710

             v₁ = 1,606 10⁴ m / s

4 0
2 years ago
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