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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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Particle 1 and particle 2 have masses of m1 = 2.2 × 10-8 kg and m2 = 4.8 × 10-8 kg, but they carry the same charge q. The two pa

Lorico [155]

Answer: $r_2=11.81 cm$

Explanation:

Given

$m_1=2.2\times 10^{-8} kg$

$m_2=4.8\times 10^{-8} kg$

same charge on both masses

potential Energy due to Magnetic Field =Kinetic Energy of Particle

$qV=\frac{mv^2}{2}$

$v=\sqrt{\frac{2qV}{m}}$

and we know

Force due to magnetic field will Provide centripetal Force

$qvB=\frac{mv^2}{r}$

$B=\frac{\sqrt{\frac{2Vm}{q}}}{r}$

and B is equal for both particles

thus $\frac{m}{r^2}=constant$

$\frac{m_1}{r_1^2}=\frac{m_2}{r_2^2}$

$r_2^2=\frac{4.8}{2.2}\times 8^2$

$r_2=11.81 cm$

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3 years ago

The moon and other satellites rotate around the earth. Identify the force that keeps these satellites in orbit. A) gravity B) fr

RoseWind [281]

The force is gravity

4 0

3 years ago

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During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 2.61

bonufazy [111]

Answer:

The shell hit at a distance of 1.9 x 10² km

The time of flight of the shell was 5.3 x 10² s

Explanation:

The position of the shell is given by the vector "r":

r = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)

where:

x0 = initial horizontal position

v0 = magnitude of the initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration of gravity

When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.

Then:

ry = 0 = y0 + v0 * t * sin α + 1/2 g t²

Since the gun is at the center of our system of reference, y0 and x0 = 0

0 = t (v0 sin α + 1/2 g t)

t= 0 is discarded as solution

v0 sin α + 1/2 g t = 0

t = -2v0 sin α / g

t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.

Then, the distance at which the shell hit is:

Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α

Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km

7 0

3 years ago

Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

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2 years ago

Which of the materials would be able to scratch Quartz

Lelechka [254]

<span>anything harder than mohs scale 7 so eg Topaz, Corundum and diamond representing mohs scale 8 9 and 10 respectively.</span>

7 0

3 years ago

Why are drills performed in drum beats ​

Why are drills performed in drum beats