All categories

2 years ago

Explain the following observation ; a boxer stands with his legs well apart

1 answer:

4 0

This stance is good for preventing punches to the face, as it keeps the opponent further away and allows the player to block with his arms. The southpaw stance is exactly the same as the upright stance, with the sole difference being that the boxer is left-handed.

They keep a wide base. What's easier to knock over, a cone or a pencil? The pencil, because it has a smaller base.

You might be interested in

How much energy (in joules) is released when 0.06 kilograms of mercury is condensed to a liquid at the same temperature?

zvonat [6]

The answer is .... C. 17,705.1 J

7 0

4 years ago

Read 2 more answers

What exactly is Density? Please answer

solmaris [256]

Answer:

density is the quantity per a unit volume, unit area or unit length.

Explanation:

7 0

3 years ago

A 2.5m long steel piano wire has a diameter of 0.5cm how great is the tention in the wire if it stretches by 0.45cm when tighten

UkoKoshka [18]

Answer: Their u go i found it their was about 3 pages i did not no what pages u had to do.

Explanation:

Download pdf

7 0

3 years ago

A potter's wheel is spinning with an initial angular velocity of 11 rad/s . It rotates through an angle of 80.0 rad in the proce

Grace [21]

The angular acceleration of the wheel approximately <u>-0.76 rad/s² or proportionally as deceleration approximately 0.76 rad/s</u>.
It need approximately <u>14.474 s</u> to come to rest.

<h2>Introduction</h2>

Hi ! I will help you to discuss about Proportionally Changes in Circular Motion. The analogy of proportionally changes in circular motion is same as the analogy of proportionally changes in direct motion. Here you will hear again the terms acceleration and change in speed, only expressed in the form of a certain angle coverage. Before that, in circular motion, it is necessary to know the following conditions:

1 rotation = 2π rad
1 rps = 2π rad/s
1 rpm = $\sf{\frac{1}{60} \: rps}$ = $\sf{\frac{1}{30}\pi \: rad/s}$

<h2>Formula Used</h2>

The following equations apply to proportionally changes circular motion:

<h3>Relationship between Angular Acceleration and Change of Angular Velocity </h3>

$\boxed{\sf{\bold{\omega_t = \omega_0 + \alpha \times t}}}$

With the following conditions:

$\sf{\omega_t}$ = final angular velocity (rad/s)
$\sf{\omega_0}$ = initial angular velocity (rad/s)
$\sf{\alpha}$ = angular acceleration (rad/s²)
t = interval of the time (s)

<h3>Relationship between Angular Acceleration and Change of $\sf{\theta}$ (Angle of Rotation) </h3>

$\boxed{\sf{\bold{\theta = \omega_0 \times t + \frac{1}{2} \times \alpha \times t^2}}}$

$\boxed{\sf{\bold{(\omega_t)^2= (\omega_0)^2 + 2 \times \alpha \times \theta}}}$

With the following condition :

$\sf{\theta}$ = change of the sudut (rad)
$\sf{\alpha}$ = angular acceleration (rad/s²)
t = interval of the time (s)
$\sf{\omega_t}$ = final angular velocity (rad/s)
$\sf{\omega_0}$ = initial angular velocity (rad/s)

<h2>Problem Solving</h2>

We know that :

$\sf{\omega_t}$ = final angular velocity = 0 rad/s >> see in the sentence "in the process of coming to rest."
$\sf{\omega_0}$ = initial angular velocity = 11 rad/s
$\sf{\theta}$ = change of the sudut = 80.0 rad

What was asked :

$\sf{\alpha}$ = angular acceleration = ... rad/s²
t = interval of the time = ... s

Step by step :

$\sf{\alpha}$ = ... rad/s²

$\sf{(\omega_t)^2= (\omega_0)^2 + 2 \times \alpha \times \theta}$

$\sf{0^2= (11)^2 + 2 \times \alpha \times 80}$

$\sf{0 = 121 + 160 \alpha}$

$\sf{-160 \alpha = 121}$

$\sf{\alpha = \frac{121}{-160}}$

$\sf{\alpha = -0.75625 \: rad/s^2 \approx \boxed{-0.76 \: rad/s^2}}$

t = ... s

$\sf{\alpha = \frac{\omega_0 - \omega_t}{t}}$

$\sf{-0.76 = \frac{0 - 11}{t}}$

$\sf{-0.76t = -11}$

$\sf{t = \frac{- 11}{-0.76}}$

$\boxed{\sf{t \approx 14.474 \: s}}$

<h3>Conclusion</h3>

So :

The angular acceleration of the wheel approximately -0.76 rad/s² or proportionally as deceleration approximately 0.76 rad/s.
It need approximately 14.474 s to come to rest.

5 0

2 years ago

Sound wave W has amplitude ‘A' and frequency ‘f'. Sound wave X is louder and lower in pitch than sound wave W. What can be said

ra1l [238]

Answer:

Sound wave X amplitude is greater than 'A' and its frequency is lesser than

'f'

Explanation:

The pitch of a sound is dictated by the frequency of the sound wave, while the loudness is dictated by the amplitude.

A high pitch sound corresponds to a high frequency and a low pitch sound corresponds to a low frequency.

The larger the amplitude of the waves, the louder the sound and vice-versa.

From the question,

Sound wave W has amplitude ‘A' and frequency 'f' and

Sound wave X is louder and lower in pitch than sound wave W.

Since sound wave X is louder, this means its amplitude is greater than 'A'.

Also, since sound wave X is lower in pitch, this means its frequency is lesser than 'f'.

7 0

3 years ago

Explain the following observation ; a boxer stands with his legs well apart​

Explain the following observation ; a boxer stands with his legs well apart