Angular Acceleration, Max/Min, and Grav. Attraction

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The discussion revolves around solving three physics homework questions related to angular acceleration, banking curves, and gravitational attraction. For the first question, the user calculated the angular acceleration of a tire using the formula for rotational motion, arriving at 40.98 radians. In the second question, they derived the range of speeds for a car on a banked curve, concluding with a speed range of 8.57 to 16.603 m/s based on given parameters. The third question involved estimating the gravitational attraction between two individuals, resulting in a force of approximately 381.11 N. The user also sought clarification on using the equation editor for formatting their post.
the7joker7
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Here are three questions on my physics homework and my attempts to solve them...am I looking at this the right way?

Question 1: A tire placed on a balancing machine in a service station starts from rest and turns through 4.7 revolutions in 1.2 seconds before reaching it's final angular speed. Calculate its angular acceleration.

My attempt: Using formula x(t) = x_{0}+ v_{0}*T + (.5)at^{2}

I got

4.7 = .5(a)(1.2^{2})

Solving for a, I got 6.522 rotations, or 40.98 radians.

Question 2: A car rounds a banked curve where the radius of curvature of the road is R, the banking angle is theta, and the coefficient of static friction is mu. Find the range of speeds the car can have without slipping up or down the road, and what is the range of speeds possible if R = 100m, theta = 10 degrees, and mu = 0.10?

The formula I pounded out was...

\sqrt{(((r*g(sin(\theta) - \mu(cos(\theta))))/(cos(\theta) + \mu(sin(\theta))} < V < \sqrt{((r*g(sin(\theta) + \mu(cos(\theta))))/(cos(\theta) - \mu(sin(\theta)))}

I plugged in the numbers and wound up with 8.57 < V < 16.603, in any case, which I'm sure is right so long as my formula is right.

Question Three: Two schoolmate, Romeo and Juliet, catch each other's eye across a crowded dance floor at a school dance. Find the order of magnitude of the gravitational attraction that Juliet exerts on Romeo and vice versa. State quanities you take as data and the values you measure or estimate for them.

I basically just guessed my own masses (Romeo is 80kg and Juliet is 70kg) and the distance between is 12m. I used the formula

((m_{1}*m_{2})/distance^{2})*gravity to get 381.11N, of magnitude 10^2~.

That work?
 
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Please post in the HW help section, and please revise you post using the text tool bar. I can't read what you wrote.

Its the equation editor in advanced mode.
 
I'm not seeing the equation editor when I went to advanced mode...where is it?
 
There is a small sigma symbol up top on the tool bar.
 
Thanks! I've cleaned it up.
 
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