Ask question

Ask question

All categories

3 years ago

10

Is it true muscle mass affects someones flexibility

1 answer:

guapka [62]3 years ago

8 0

It is very very true

You might be interested in

An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e

White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

$x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)$

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

$\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}$

with this value you can compute the frequency:

a)

$f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz$

b)

the mass of the block is given by the formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg$

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

$\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s$

Finally, the amplitude is:

$x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m$

5 0

2 years ago

The observation deck of a skyscraper is 420 m above

Reil [10]

2e min :)) pls park braliest

5 0

2 years ago

Which best describes fusion nuclear reactors? They use rods containing uranium for fuel. They are currently used for electricity

Ludmilka [50]

They has been very successful but they are very expensive to operate that is your answer I hope this helps

6 0

3 years ago

Read 2 more answers

Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi

Troyanec [42]

Answer:

The lowest possible frequency of sound is 971.4 Hz.

Explanation:

Given that,

Distance between loudspeakers = 2.00 m

Height = 5.50 m

Sound speed = 340 m/s

We need to calculate the distance

Using Pythagorean theorem

$AC^2=AB^2+BC^2$

$AC^2=2.00^2+5.50^2$

$AC=\sqrt{(2.00^2+5.50^2)}$

$AC=5.85\ m$

We need to calculate the path difference

Using formula of path difference

$\Delta x=AC-BC$

Put the value into the formula

$\Delta x=5.85-5.50$

$\Delta x=0.35\ m$

We need to calculate the lowest possible frequency of sound

Using formula of frequency

$f=\dfrac{nv}{\Delta x}$

Put the value into the formula

$f=\dfrac{1\times340}{0.35}$

$f=971.4\ Hz$

Hence, The lowest possible frequency of sound is 971.4 Hz.

8 0

3 years ago

A 6.2 kg cannonball is fires from a 280 kg cannon. The cannon recoils with a velocity of 4.9 m/s. What is the velocity of the ca

lara [203]

Answer:

666 bc its on its way to my house

Explanation:

3 0

3 years ago