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# Tutor profile: Ian F.

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Ian F.
Chemist at UC Davis
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## Questions

### Subject:Chemistry

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Question:

What is the pH of .1M HCL.

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Ian F.

Take the negative log of .1 giving you 1.

### Subject:Calculus

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Question:

What is the derivative of f(x)=x^2+3/2x^3/2. And what is the partial directive in respect to x and y for y=3x+4y^2.

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Ian F.

f'(x)= 2x+9/4x^(1/2) using power rule. dy/dx is 3/(1-8y) using power rule and then solving for dy/dx the same can be done for dx/dy or just taking the inverse of dy/dx. To take partial directives you take the derivative in respect to what is on the bottom of dy/dx being x or dx/dy being y. So that when you do it what you take the derisive in respect to you do the normal derisive of but if it doesn't contain that variable then you take it normally but add dx/dy or dy/dx respectively to the end.

### Subject:Physics

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Question:

What height velocity and angle is a baseball that is thrown 30 m/s at 25 degrees at time equals 2 seconds. What is the balls maximum height in the y direction. What is the balls maximum distances in the x direction. How long does it take for the ball to get to its maximum height in the y direction. The ball weighs 2 kg.

Inactive
Ian F.

The first thing I would do is make a motion map and force diagram chart so that you know what is happening to the ball as time goes on; this will be helpful in general. First we need to find the x and y values of the initial velocities. For the Y value of the initial velocity take sin(25)*30m/s= 12.678 m/s, for the X value take the cos(25)*30m/s= 27.189m/s. To figure out how long it takes to get to max height use Vfinal=Vinital-at; 0=12.678m/s*T and solve for T, T=1.29 sec. You use the velocity in the y direction because that is the only component that will contribute to an increase in the y value, since all other components dont have a y value. For maximum x direction or rather where is it going to land multiple the time it took to get to the max y value multiple it by 2 and then multiple that by the x value velocity; T=1.29sec*2=2.58 sec; x=vt=(27.189m/s)*2.58=70.15m. For the maximum y value height you will use Vfinal^2=Vinital^(2)+2a(xfinal-xinital) and solve for x by setting Vfinal equal to zero since at the max heigh the ball will not be moving. 0=12.678^(2)+2(-9.8m/s^2)(x-0), Xfinal=8.2m. To solve for the angle and velocity at 2 seconds you need to find the y and x values of the velocity at that time period. You solve for each of these separately and then calculate the total velocity and angle. The velocity in the x direction doesn't change since their is no force acting against it in that direction. For the y direction gravity is acting against it so solve for the final velocity at time equals 2 seconds. Vfinal=Vinital+aT; Vfinal=12.678+(-9.8m/s^2)*2sec=-6.922m/s. So since we know that at 2 seconds the velocity in the y direction is negative you can now draw a triangle that shows those components in quadrant 4 with a positive 27.189 x component and a negative 6.922 value in the Y direction. To solve for the total velocity take the square root of the sum squares Vfinal^2=Vx^(2)+Vy^(2); Vfianl^2=(27.289^2)+(6.922^2)= 28.05 m/s. To solve for the angle take the arctan of 6.922/27.189 to get 14.28 degrees.

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