# Tutor profile: Robert M.

## Questions

### Subject: Electrical Engineering

If an synchronous electrical motor has a supplied frequency of $$50Hz$$ and $$12$$ $$poles$$. What is the motor speed of the motor, and how long would it take for the motor to complete $$10^6$$ $$revolutions$$ in days?

First, list known values: $$f=50Hz$$ $$P=12Poles$$ $$\omega=10^6$$ The motor speed can be found, which is measured in revolutions per minute $$(rpm)$$: $$\omega_{motor}=\frac{2f}{P}$$ $$\omega_{motor}=\frac{2(50)}{(12)}$$ $$\omega_{motor}=8.\overline3rps$$ $$\omega_{motor}=500rpm$$ The time can now be found: $$t=\frac{\omega}{\omega_{motor}}$$ $$t=\frac{(10^6revolutions)}{500rpm}$$ $$t=2000min$$ $$t=1.3\overline8days$$

### Subject: Algebra

Consider the following two equations: $$3x+12y=81$$ and $$15x+22y=25$$ Find the values for $$x$$ and $$y$$ using substitution.

First, take the first equation $$3x+12y=81$$ and rearrange for $$x$$. $$3x+12y=81$$ $$3x=81-12y$$ $$x=\frac{81-12y}{3}$$ $$x=27-4y$$ Second, substitute $$x=27-4y$$ into the second equation $$15x+22y=25$$, then rearrange to find the value of $$y$$. $$15x+22y=25$$ $$15(27-4y)+22y=25$$ $$405-60y+22y=25$$ $$-38y=25-405$$ $$-38y=-380$$ $$y=\frac{-380}{-38}$$ $$y=10$$ Now that $$y$$ has been found, substitute this value into the first equation $$3x+12y=81$$, then the value of $$x$$ can be found. $$3x+12y=81$$ $$3x+12(10)=81$$ $$3x+120=81$$ $$3x=81-120$$ $$3x=-39$$ $$x=\frac{-39}{3}$$ $$x=-13$$ $$\therefore$$ $$x=-13$$ and $$y=10$$.

### Subject: Physics

A ball with an external radius of $$150 mm$$ is dropped into the center of a vertical cylinder with an internal radius of $$1.1 m$$. The ball has a horizontal movement of $$0.12km/h$$ at the time of release. Assuming: $$*$$ Gravity $$=9.8m/s^2$$. $$*$$ No air resistance/no terminal velocity. $$*$$ The cylinder does not have an end. How long will it take for the ball to touch the internal face of the cylinder in seconds? At this time, how far would the ball have vertically travelled?

The first step is to list out the known values that have been presented in the question and assumptions made. This is done in the units displayed. Knowns: $$r_{ball}=150mm$$ $$r_{cylinder}=1.1m$$ $$v_{horizontal}=0.12km/hr$$ $$g=9.8m/s^2$$ Now that the known values have been listed out, they will all be converted to SI. Knowns Converted: $$r_{ball}=150mm=\frac{150}{1000}m=0.15m$$ $$r_{cylinder}=1.1m$$ $$v_{horizontal}=0.12km/hr=\frac{0.12*1000}{60^2}m/s=\frac{1}{30}m/s$$ $$g=9.8m/s^2$$ The unknowns will now be displayed: $$t=?s$$ $$h=?m$$ Firstly, the time will be found: $$t=\frac{\Delta r}{v_{horizontal}}$$ $$t=\frac{(r_{cylinder}-r_{ball})}{v_{horizontal}}$$ $$t=\frac{(1.1m)-(0.15m)}{(\frac{1}{30}m/s)}$$ $$t=\frac{0.95m}{\frac{1}{30}m/s}$$ $$t=\frac{0.95m*30s}{1m}$$ $$t=0.95*30s$$ $$t=28.5s$$ Secondly, the vertical height will be found: $$h=\frac{g*t^2}{2}$$ $$h=\frac{(9.8m/s^2)*(28.5s)^2}{2}$$ $$h=\frac{9.8m/s^2*812.25s^2}{2}$$ $$h=\frac{7960.05m}{2}$$ $$h=3980.025m$$ $$\therefore$$ it will take the ball $$28.5s$$ to touch the internal face of the cylinder, and it would have travelled $$3980.025m$$ vertically at this time.

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