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galben [10]
3 years ago
9

A tornado moving cross a street with a force of 600 N pushes a trailer a distance of 25 m.

Physics
1 answer:
stealth61 [152]3 years ago
3 0

Answer:

<em>W = 15,000 Joule</em>

Explanation:

<u>Mechanical Work </u>

Mechanical work is the amount of energy transferred by a force. It's a scalar quantity, with SI units of joules.

If both the force and displacement are parallel, then we can use the scalar formula:

W=F.s

Where

F =Magnitude of the force

s = magnitude of the displacement

The tornado moved a trailer with a force of F=600 N for a distance of s=25 m. The work done was:

W = 600 N * 25 m

W = 15,000 Joule

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Suppose you are in an elevator. As the elevator starts upward, its speed will increase. During this time when the elevator is mo
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Answer:increased

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4 years ago
Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f
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Answer:

\large \boxed{\text{30 rev/s}}

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

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ω₁ = 6.0    rev·s⁻¹

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Calculation:

\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}

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4 years ago
A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit
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Answer:

The magnitude of angular acceleration is 232.38\ rad/s^2.

Explanation:

Given that,

Initial angular velocity, \omega_i=3500\ rev/min=366.5\ rad/s

When it switched off, it comes o rest, \omega_f=0

Number of revolution, \theta=46=289.02\ rad

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2  

So, the magnitude of angular acceleration is 232.38\ rad/s^2. Hence, this is the required solution.

6 0
3 years ago
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