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2 years ago

Why are drills performed in drum beats

Physics

2 answers:

Whitepunk [10]2 years ago

6 0

Answer:

The drum drill is just one option in stride frequency development. Most of the time, the drum drill can be seen as just a rhythm drill that allows an athlete to relax and experiment with the right range of motion and bounce. A solid background in floating drills and developing reactivity should help athletes mold their stride into a balanced motion that maximizes their speed.

I have used frequency drills for years and now understand the nature of stride development mainly from shaping the stride parameters we all have known about for a long time. The drum drill is a special exercise that can make a great change in athletes who are receptive to improving and with a coach who is worth their salt in instruction. The drum drill is just one option for improving an athlete, and it’s more than fine to use any method you see fit that helps improve stride frequency.

Explanation:

PtichkaEL [24]2 years ago

6 0

Explanation:

Most of the time, the drum drill can be seen as just a rhythm drill that allows an athlete to relax and experiment with the right range of motion and bounce. A solid background in floating drills and developing reactivity should help athletes mold their stride into a balanced motion that maximizes their speed

We break down a UK drill beat, a genre that rapidly became one of the ... Beat Dissected is a regular series in which we deconstruct drum patterns, ... an 808 bass line with plenty of glides and a piano played back in halftime.

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Which of the following ways of writing 1000w is incorrect?

xenn [34]

Answer:

the third one is incorrect

Explanation:

10 x 10³= 10^1 x 10^3 = 10^4

8 0

2 years ago

After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48

masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

$15min*\frac{1hr}{60min}=0.25hr$

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

$V=\frac{distance}{time}$

the plane traveled a distance to the north of 48km so the velocity is:

$V=\frac{48km}{0.25hr}$

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

$V_{x}=42km/hr*cos(-19^{o} )=39.71 i$

and

$V_{y}=42km/hr*sin(-19^{o} )=-13.67 j$

So the velocity of the wind can be expressed as a vector as:

$V_{wind}=(39.71i - 13.67j)km/hr$

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

$V_{plane x}=V_{plane/wind x}+V_{wind x}$

$V_{plane/wind x}=V_{plane x}-V_{wind x}$

$V_{plane/wind x}=(0-39.71 i)km/hr$

$V_{plane/wind x}= -39.71 i km/hr$

and

$V_{plane y}=V_{plane/wind y}+V_{wind y}$

$V_{plane/wind y}=V_{plane y}-V_{wind y}$

$V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr$

$V_{plane/wind x}= 205.67 j km/hr$

So the velocity of the plane with respect to the wind can be rewritten as:

$V_{plane/wind x}= (-39.71i + 205.67 j) km/hr$

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

$speed=\sqrt{(-39.71)^{2}+(205.67)^{2} }$