/sci/ - Science & Math

>>11410383
>why is calculus so retarded?
Because it isn't real math. It's just a really old fashioned thing they teach engineers and scientists to give them some basic tools to deal with observed laws of nature.
>infinitesimals
For your intents and purposes, [math] \frac{\text{d}y}{\text{d}x} [/math] is the ratio between two very small (infinitesimal) quantities.
>U substitution
A method of changing the variable of integration that aims to make it easier to find the anti-derivative of the integrand. If you've got [math] \int f(x)\ \text{d}x [/math] but f(x) is too complicated a function to anti-differentiate, then sometimes you can let [math] u=g(x) [/math] which implies [math] \text{d}u=g'(x) \text{d} x [/math]. If you picked a good u then [math] \int f(x)\ \text{d} x=\int f(g(x))g'(x)\ \text{d}x=\int f(u)\ \text{d}u [/math] is much easier to deal with.
>but couldnt you just turn the large gear fewer radians
Sometimes you want to multiply the input force (torque, really) and sometimes you want to multiply the input angular speed. Say you are imputing some force to drive gear 1 with radius R1. That gear then drives gear two with radius R2. If you are more conserned about forces, then the transfer function (ratio of output to input) of the system is [math] \frac{T_2}{T_1}=\frac{R_2}{R_1} [/math]. So if R2 is greater than R1, you can literally multiply the amount of torque you put into the system. But if you are more concerned with speed, then the transfer function is [math] \frac{\omega_2}{\omega_1}=\frac{R_1}{R_2} [/math]. So if R1 is greater than R2, you can multiply the input speed (at the cost of torque). Sometimes you want one setup, sometimes you want the other. It depends on application. These equations are very easy to derive and I recommend trying it out for yourself if you are still confused.
>>11410437
>>11410453
How was that?

>>11364594
>The cross-sectional area is split into 16 smaller squares, we've measured each square, then derived an average.
Perfect. The average of all those velocities is the average airspeed in the duct. Then you have it that (mass flow rate) = (average airspeed)*(area of cross section)*(density). This is the most accurate method, I'd say.
>I just used the two co-efficient equation and got a value very close to a table
wonderful~
>>11364599
better than (((Kelvin)))

>>11325309
>>11325163
I can't be arsed to go through your calculation, but the most important thing to realize about this problem is (1) the acceleration of each block is the same and (2) the linear acceleration of the blocks is just the radius of the pulley times its angular acceleration.
First, sum the torques on the pulley:
[eqn] \sum\tau=I\dot\omega\implies T_1-T_3+\frac{1}{2}m_pa=0 [/eqn]
Then, block 1:
[eqn]\sum F=m_1a\implies T_1-m_1a=m_1g(\sin\theta-\mu\cos\theta)[/eqn]
Then, block 3:
[eqn] \sum F=m_3a\implies T_3+m_3a=m_3g [/eqn]
So you have three linear equations with three unknowns. You may easily solve for both tensions and acceleration at this point.
[eqn] a=\frac{m_3g+m_1g(\mu\cos\theta-\sin\theta)}{m_1+m_3+m_p/2} [/eqn]
>>11325298
When you hit the thing with a hammer, there is an extreme concentration of stress right around where the hammer impacts, much much greater than if you where to stand or even jump. Stress concentrations are the enemy of built-up components, like a few two by fours fastened with some nails. Also, the hammer is putting a lot more than 12 oz of force on the thing (something closer to about 100 lbs, from google) depending on how fast you swing it, how hard the wood is, etc.
>>11325390
>pours
Anon, your hands dry because the water evaporates, not because it is going into your blood stream. What the fuck
>>11325517
>>11325522
good boy~

>>11325309
>>11325163
I can't be arsed to go through your calculation, but the most important thing to realize about this problem is (1) the acceleration of each block is the same and (2) the linear acceleration of the blocks is just the radius of the pulley times its angular acceleration.
First, sum the torques on the pulley:
[eqn] \sum\tau=I\dot\omega\implies T_1-T_3+\frac{1}{2}m_pa=0 [/eqn]
Then, block 1:
[eqn]\sum F=m_1a\implies T_1-m_1a=m_1g(\sin\theta-\mu\cos\theta)[/eqn]
Then, block 3:
[eqn] \sum F=m_3a\implies T_3+m_3a=m_3g [/eqn]
So you have three linear equations three unknowns. You may easily solve for both tensions and acceleration at this point.
[eqn] a=\frac{m_3g+m_1g(\mu\cos\theta-\sin\theta)}{m_1+m_3+m_p/2} [/eqn]
>>11325298
When you hit the thing with a hammer, there is an extreme concentration of stress right around where the hammer impacts, much much greater than if you where to stand or even jump. Stress concentrations are the enemy of built-up components, like a few two by fours fastened with some nails. Also, the hammer is putting a lot more than 12 oz of force on the thing (something closer to about 100 lbs, from google) depending on how fast you swing it, how hard the wood is, etc.
>>11325390
>pours
>>11325517
good boy