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# Tutor profile: Katya L.

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Katya L.
AP Physics Teaching Assistant
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## Questions

### Subject:Physics (Newtonian Mechanics)

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Question:

A car of mass M is moving through a flat curved section of highway at 50m/s. The coefficient of static friction between the tire and the roadway is .5. What is the radius of curvature of the highway?

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Katya L.

Since the car is turning it is experiencing centripetal acceleration. The only force acting on the car in the x direction, parallel to ground, is the force of friction so therefore the centripetal acceleration is due to friction. We can write this as $$\sum F_{x}=F_{f}=ma_{c}=m\frac{v^2}{r}$$ equation 1 where $$\frac{v^2}{r}$$is the definition of centripetal acceleration $$a_{c}$$. Using the definition of the force of static friction we can write $$F_{f}=\mu_{s}N=m\frac{v^2}{r}$$ equation 2 where $$u_{s}$$ is the coefficient of static friction and N is the normal force. To solve for Normal force we can write an equation for forces on the car in the y-direction, perpendicular to the ground so that $$\sum F_{y}=N-F_{g}=ma$$ equation 3 where $$F_{g}$$ is the force due to gravity which is defined as mg. Since the car is not accelerating in the y direction we can set equation 3 equal to 0. Using the definition of force due to gravity we can write $$\sum F_{y}=N-mg=0$$ equation 4 $$N=mg$$ equation 5 Using equation 5 we can rewrite equation 2 such that $$F_{f}=\mu_{s}mg=m\frac{v^2}{r}$$ equation 6 dividing both sides by m we have $$\mu_{s}g=\frac{v^2}{r}$$ equation 7 solving for r we have $$r=\frac{v^2}{\mu_{s}g}$$ equation 8 Finally plugging in our given values (assuming $$g=10m/s^2$$) we have $$r=\frac{{50m/s}^2}{.5(10m/s^2)}=50m$$

### Subject:Algebra

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Question:

5+3(x−1)=3x−2(x−3) Solve for x.

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Katya L.

First, we distribute the 3(x-1) on the left side and the -2(x-3) on the right side. Distributing means you multiply the first term by all the terms in the parentheses. For example on the left side, you multiply both the x by 3 and the -1 by 3 so that 3(x-1)=3x-3 performing the distribution of the right side we then have -2(x-3)=-2x+6 This gives us the new equation 5+3x-3=3x-2x+6 Next, we add like terms on each side. So on the left side, we add the 5 and -3 together and on the right side, we subtract 2x from 3x. This gives us 2+3x=x+6 Subtracting x from both sides we have 2+3x-x=6 and subtracting 2 from both sides we have 3x-x=6-2 combing each side we have 2x=4 Finally dividing each side by 2 we have x=2 We can check this by substituting 2 in for x into the original equation so 5+3(2-1)=3(2)-2(2-3) 5+3(1)=6-2(-1) 8=8 Since both sides equal each other this is a good indicator that out the answer is correct.

### Subject:Physics

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Question:

A muon has a proper lifetime of 2.2μs. If the muon as a speed of 99% the speed of light, what is the average distance it travels before decaying?

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Katya L.

The proper lifetime is another way of saying the lifetime in the frame of the particle. Because the muon is traveling close to the speed of light then to find the lifetime of the muon in the frame of the observer, or the laboratory frame, it is equal to the proper time multiplied by the lorentz factor, γ. The definition of the lorentz factor is $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ Therefore, denoting $$\tau_{0}$$ as the proper lifetime and t as the lifetime in the lab frame we have $$t=\gamma\tau_{0}= \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\tau_{0}=\frac{1}{\sqrt{1-\frac{0.99c^2}{c^2}}}2.2μs$$ We can then find the distance it travels before decaying since the definition of velocity is distance over time, the distance, d, will be the velocity multiplied by the time in the lab frame so $$d=vt=0.99c\frac{1}{\sqrt{1-\frac{0.99c^2}{c^2}}}2.2μs=4.6x10^{3}m$$

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