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Carter K.
Aerospace Engineering Student at the University of Washington
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Physics (Newtonian Mechanics)
TutorMe
Question:

A ball moves around a banked, circular track with a constant velocity of 2 m/s. The radius of the circular track is 0.5 m. Does the ball have an acceleration? If so, what is the value of this acceleration?

Carter K.
Answer:

Although the ball is moving with constant velocity, it does have a non-zero acceleration. Recall that velocity is a vector quantity. While the magnitude of the velocity is not changing, the direction of the vector is, and therefore the acceleration must be non-zero. For constant rotational motion, this acceleration is called centripetal acceleration. The direction of this acceleration is radially inwards towards the center of the circle. Calculation of $$a_r$$: $$a_r = \frac{v^2}{r} = \frac{2^2}{0.5} = 8 \frac{m}{s^2}$$

Calculus
TutorMe
Question:

The change in volume of a tub of water is given by the function $$f(t) = 4t + 2$$ $$(\frac{m^3}{s})$$. Calculate the mass of water in the tub after 30 seconds. Use a density of 1 $$\frac{kg}{m^3}$$ for water.

Carter K.
Answer:

The mass of the water can be calculated by multiplying the density times volume. Therefore, the change in mass is given by $$\rho*f(t)$$. To get the total change in mass, integrate this expression from 0 to 30 seconds. $$\int_{0}^{30} (1) * (4t +2) dt = (2t^2 + 2t) \Big|_0^{30} = [2*(30^2) + 2*30] - [2*(0^2) + 2*0]$$ $$\Rightarrow 2*900 + 60 = 1860$$ Therefore, the total mass of water in the tub after 30 seconds is 1860 kg

Algebra
TutorMe
Question:

A ball is thrown straight up from an initial height of zero meters with an initial velocity of 3 m/s. After what time will it hit the ground? Use g = 9.8 m/s^2

Carter K.
Answer:

Model the flight of the ball using a quadratic equation. $$-\frac{1}{2}*g*t^2 + v_0*t + x_0 = \Delta y = 0$$ $$v_0 = 3$$ m/s $$x_0 = 0$$ m $$-\frac{1}{2}*9.8*t^2 + 3t = 0$$ $$t*(-4.9t + 3) = 0$$ Solving for t, we get: $$t = 0$$ s, $$t = 0.61$$ s $$\Rightarrow$$ The ball is in the air for $$0.61$$ s

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