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3 years ago

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A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it

that extends from the axis of the disk to the rim. At t = 0 this line lies along the x-axis and the disk is rotating with positive angular velocity ω0z. The disk has constant positive angular acceleration αz. At what time after t = 0 has the line on the disk rotated through an angle θ?
Express your answer in terms of the variables ω0z, αz, and θ.

1 answer:

natka813 [3]3 years ago

8 0

Answer:

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Explanation:

The rotated angle is given by:

$\theta=\omega0*t+1/2*\alpha*t^2$

Since this is a quadratic equation it can be solved using:

$x=\frac{-b \± \sqrt{b^2-4*a*c} }{2*a}$

Rewriting our equation:

$1/2*\alpha*t^2+\omega0*t-\theta=0$

$t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Since $\sqrt{\omega0^2+2*\theta*\alpha} >\omega0$ we discard the negative solution.

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

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3 years ago

A 99.5 N grocery cart is pushed 12.9 m along an aisle by a shopper who exerts a constant horizontal force of 34.6 N. The acceler

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1) 9.4 m/s

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where

F = 34.6 N is the magnitude of the force

d = 12.9 m is the displacement of the cart

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W = (34.6 N)(12.9 m) = 446.3 J

According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:

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Since the cart was initially at rest, $K_i = 0$ , so

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v is the final speed

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$m=\frac{F_g}{g}=\frac{99.5 N}{9.8 m/s^2}=10.2 kg$

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W = Fd

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d is the displacement of the train

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$W=(4.28\cdot 10^5 N)(586 m)=2.51\cdot 10^7 J$

3) $2.51\cdot 10^7 J$

According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:

$W=\Delta K = K_f - K_i$

where

W is the work done

$\Delta K$ is the change in kinetic energy

Therefore, the change in kinetic energy is

$\Delta K = W = 2.51\cdot 10^7 J$

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where

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Re-writing the equation,

$W=K_f = \frac{1}{2}mv^2$

where

m = 36300 kg is the mass of the train

v is the final speed of the train

Solving for v, we find

$v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(2.51\cdot 10^7 J)}{36300 kg}}=37.2 m/s$

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4 years ago

You put a 3 kg block in the box, so the total mass is now 9 kg, and you launch this heavier box with an initial speed of 5 m/s.

OlgaM077 [116]

Answer:

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Explanation:

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3 years ago