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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.

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
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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.

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
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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.

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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