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3 years ago

What are 5 violations of the rules for Ultimate Frisbee

Physics

1 answer:

aniked [119]3 years ago

4 0

Answer:

Violations. A violation occurs when a player violates the rules in a manner which does not result in physical contact ( e.g. throwing a pass during an approach to the goal line; illegal guarding position by the marker; not establishing a pivot foot after carrying the disc in from out-of-bounds, etc.)

The object of Ultimate Frisbee is to gain points by scoring goals. The disc may only be passed, and a goal is scored when a player successfully passes the disc to a teammate in the end zone which that team is attacking. The team with the most points at the end of the game is declared the winner.

Explanation:

Vert (Vertical) Stack – The most common offensive play in Ultimate Frisbee. The Mailman – An intense offensive tactical maneuver to guarantee a gain of at least 70% of the field. TurboDisc – Can be added to almost any play to ensure max effectiveness.

If contact occurs and the thrower drops the disc, the count stops until the thrower picks up the disc and continues with play. The defensive player will have to back up 5 yards (15 feet) to allow a free throw from the offensive player.

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power of a crane is 25000 watt calculate the time required by it to lift a load of 6000 kg up tp the height of 20 m

saul85 [17]

Solution:

We have,

Power [P] = 25000 Watt

Mass [m] = 6000 kg

Height [h] = 20 metres

Time [t] = ?

Now,

P = W/t = F x d/t = mxgx h/t

Or, 25000 = 6000 x 10 x 20/25000 [.......g = 10

m/s^2]

Or, t = 6000 x 10 x 20/25000

Or, t = 1200/25

Therefore, t = 48 second

Hence, the required time for the crane to lift the load is 48 seconds.

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3 years ago

How does half-life apply to real life?

wlad13 [49]

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3 years ago

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poizon [28]

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3 years ago

How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the

kirza4 [7]

Answer:

Approximately $6.2\; {\rm rpm}$ , assuming that the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ .

Explanation:

Let $\omega$ denote the required angular velocity of this Ferris wheel. Let $m$ denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

Weight of the passenger (downwards), $m\, g$ , and possibly
Normal force $F_\text{normal}$ that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is $0$ - that is, $F_\text{normal} = 0$ .

The net force on this passenger is $(m\, g - F_\text{normal})$ . Hence, when $F_\text{normal} = 0$ , the net force on this passenger would be equal to $m\, g$ .

Passengers on this Ferris wheel are in a centripetal motion of angular velocity $\omega$ around a circle of radius $r$ . Thus, the centripetal acceleration of these passengers would be $a = \omega^{2}\, r$ . The net force on a passenger of mass $m$ would be $m\, a = m\, \omega^{2}\, r$ .

Notice that $m\, \omega^{2} \, r = (\text{Net Force}) = m\, g$ . Solve this equation for $\omega$ , the angular speed of this Ferris wheel. Since $g = 9.81\; {\rm m\cdot s^{-2}}$ and $r = 23\; {\rm m}$ :

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$ .

$\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}$ .

The question is asking for the angular velocity of this Ferris wheel in the unit ${\rm rpm}$ , where $1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})$ . Apply unit conversion:

$\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}$ .

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2 years ago

Using information about natural laws, explain why some car crashes produce minor injuries and others produce catastrophic injuri

kicyunya [14]

The forces in some car crashes are greater than others; therefore, the body can be more severely injured by greater and more powerful forces. I hope this helped! :^)

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3 years ago