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zepelin [54]
3 years ago
14

WA

Physics
1 answer:
brilliants [131]3 years ago
3 0

Answer:

the answers the correct one is d

Explanation:

The speed of sound is constant so we can use the relations of uniform motion

           v = x / t

            x = v t

now let's calculate the distance for each person

t = 5s

         x₁ = 300 5

         x₁ = 1500 m

t = 6s

         x₂ = 300 6

         x₂ = 1800 m

therefore we have two possibilities

a) the two people are on the same side, therefore the distance between them is

         Δx = x₂- x₁

         Δx = 1800 - 1500

         Δx = 300 m

       

let's reduce to km

         Δx = 0.300 km

b) people are on opposite sides of the sound

         Δx = x₂ + x₁

         Δx = 1800 + 1500

         Δx = 3300 m

         Δx = 3.3 km

when checking the answers the correct one is d

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Answer:

<h2>0.5 m/s²</h2>

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a =  \frac{f}{m}  \\

From the question we have

a =  \frac{5}{10} =  \frac{1}{2}   \\  = 0.5

We have the final answer as

<h3>0.5 m/s²</h3>

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2 years ago
A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result
vivado [14]

Answer:

21.6\ \text{kg m}^2

3.672\ \text{Nm}

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Explanation:

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\omega_f = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2

Torque is given by

\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2

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t = 60.6 s

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\omega_f = 0

\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2

Frictional torque is given by

\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}

The magnitude of the torque caused by friction is 3.672\ \text{Nm}

Speeding up

\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}

Slowing down

\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}

Total number of revolutions

\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}

\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}

The total number of revolutions the wheel goes through is 54.66\ \text{revolutions}.

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