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# Tutor profile: Anthony A.

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Anthony A.
Recent Engineering Grad w/ Tutoring Experience
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## Questions

### Subject:Calculus

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Question:

You are blowing up a spherical balloon at at rate of 200 cm^3/s. When the balloon has a diameter of 10 cm, how fast is its diameter expanding?

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Anthony A.

This is a related rates problem. In this case, we know what the rate of filling the balloon is (200 cm^3/s) and this is dV/dt, or the time-differential change in volume. Since the balloon is a sphere, an equation relating the volume of a sphere to its diameter would be helpful. We have the equation... V = (4/3)*pi*(W/2)^3 Where V is volume and W is the diameter. Taking the derivative with respect to time of both sides of the equation, we get... d/dt *V = (4/3)*pi*(1/8)*W^3*d/dt => dV/dt = (1/2)*pi*W^2*dW/dt dV/dt is given to be 200 and W, at the time the question specifies, is 10. Re-arranging the equation and solving for dW/dt... (200)*(2)/(pi*10^2) = dW/dt = 1.27 cm/s

### Subject:Electrical Engineering

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Question:

An inverting op-amp is configured so that a 10k resistor (R1) is connected to the negative terminal and a 10k resistor (R2) is connected from the negative terminal to the output. Furthermore, a 0.1uF (C1) capacitor is connected in parallel with the second resistor. The input sinusoid signal is at the positive terminal and has a frequency of 1000 rad/s. What is the overall gain of this amplifier configuration?

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Anthony A.

The gain equation for an inverting amplifier is easy enough... A = -R2/R1 However, the resistance R2 in this case is no simply the resistor value, because of the capacitor. We have to find the parallel combination of R2 and C1, or 10k | | (1/jwC1) where j is an imaginary number and w is frequency. The parallel combination equation becomes... [10k*(1/jwC1)] / [10k + (1/jwC1)] => 10k/(10k*jwC1 +1) Substituting values offers the following... 10k/(1+j) Using properties of complex division, the result is... 10000/sqrt(2) = 7.07k < 45 degrees = R2 This is the new R2 value. Gain, therefore, would be... A = -7.07k/10k = -.707

### Subject:Physics

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Question:

A golf ball is struck, from level ground, at an initial velocity of 25 m/s and an angle of 40 degrees. Neglecting resistance of air, for how long will the golf ball be in the air?

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Anthony A.

When we want to know how long a projectile stays in the air, we only really compare about the part (or component) of velocity that fights gravity. This is the y-component of velocity. To find the Y-component of velocity, we need to use trigonometry… V_y=(25 m/s)*sin⁡(40)=16 m/s This is the part of velocity that goes up and down. We could just as easily find the X-component by using cosine. That would help us find how far the golf ball would go. Since we just want to know how long the ball is in the air, we need to find how long a projectile traveling upward at 16 m/s would stay in the air. To find this, we use the projectile motion equation… ΔY=Δt⋅V_y - 1/2⋅a_g⋅Δt^2 Where ΔY is the change in distance (zero, since we want to end where we started, at ground level). V is the velocity’s Y-component, a_g is the acceleration due to gravity, and Δt is the change in time for which we are solving. Rearranging the projectile motion equation gives… Δt(1/2 a_g Δt - V_y )=0 When gravity is 9.81 m/s^2 (a constant) and V_y is 16 m/s (previously solved), we get the following… Δt(4.905Δt - 16)=0 Therefore, we can conclude that two solutions for time, zero (because the ball starting at the ground) and 3.26 seconds. This second solution is the final solution, as it the time it takes for the ball to return to the ground after being struck.

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