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3 years ago

Do you get more flexible by stretching

Physics

2 answers:

deff fn [24]3 years ago

8 0

Answer:

Well yes you can get more flexible by stretching

Explanation:

Altho it depends on what you eat and were you live in the world different stretches can help or hurt your body, for more information you may want to look up stretching results and practices.

Alex Ar [27]3 years ago

3 0

Answer:

Yes

Explanation:

By stretching, parts pf your body loosen and become more flexible.

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A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan

Nesterboy [21]

Answer:

Compression distance: $d \approx 0.102\,m$

Explanation:

According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy ( $K$ ), in joules, at the expense of elastic potential energy ( $U$ ), in joules, and overcoming work losses due to friction ( $W_{l}$ ), in joules:

$K + W_{l} = U$ (1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$ (2)

Where:

$m$ - Mass of the block, in kilograms.

$v$ - Final velocity of the block, in meters per second.

$\mu$ - KInetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$s$ - Width of the rough patch, in meters.

$k$ - Spring constant, in newtons per meter.

$d$ - Compression distance, in meters.

If we know that $m = 1.2\,kg$ , $v = 2.3\,\frac{m}{s}$ , $\mu = 0.44$ , $g = 9.807\,\frac{m}{s^{2}}$ , $s = 0.05\,m$ and $k = 730\,\frac{N}{m}$ , then the compression distance of the spring is:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$

$m\cdot v^{2} + 2\cdot m\cdot g \cdot s = k\cdot d^{2}$

$d = \sqrt{\frac{m\cdot (v^{2}+2\cdot g\cdot s)}{k} }$

$d \approx 0.102\,m$

4 0

3 years ago

A student is experimenting with some insulated copper wire and a power supply. She winds a single layer of the wire on a tube wi

OverLord2011 [107]

Answer:

$P=214.7187\,W$

Explanation:

Given that:

Diameter of the solenoid, $D=10\,cm=0.1\,m$

length of the solenoid, $L=90\,cm=0.9\,m$

diameter of the wire, $d=0.1\,cm=10^{-3}\,m$

magnetic field at the center of the solenoid, $B=7.4\times 10^{-3}\,T$

Now we need the no. of turns incorporated in the length of 90 cm:

$N=\frac{Length\,\,of\,\,solenoid}{diameter\,\,of\,\, wire}$

$N=\frac{L}{d}$

$N=\frac{0.9}{10^{-3}}$

$N=900\,\,turns$

For solenoids we have:

$B=\mu.n.I$ ...............................(1)

where:

$\mu=$ permeability of the medium

n = no. of turns per unit length

I = current in the coil

So,

$n=\frac{900}{0.9}$

$n=1000\,turns\,.\,m^{-1}$

Now putting the respective values in the eq. (1)

$7.4\times 10^{-3}=4\pi\times10^{-7}\times 1000\times I$

$I=5.8887\,A$

For copper we have resistivity:

$\rho=1.72\times 10^{-8}\, \Omega.m$

We know that resistance is given by:

$R=\rho.\frac{l}{a}$ .....................................(2)

where:

l = length of the conducting wire

a = cross sectional area of the conducting wire

Now we need the length (l) of the wire:

Circumference of the solenoid,

$C=\pi.D$

$C=0.1\pi\,m$

$\therefore l=C\times N$

$l=90\pi\,m$

Cross-sectional area of wire:

$a=\pi.\frac{d^2}{4}$

$a=\pi. \frac{(10^{-3})^2}{4}\,m^2$

Resistance from eq. (2):

$R=1.72\times 10^{-8}\times \frac{90\pi}{\pi. \frac{(10^{-3})^2}{4}}$

$R=6.192 \,\Omega$

For power we have:

$P=I^2.R$

$P=5.8887^2 \times 6.192$

$P=214.7187\,W$

6 0

3 years ago

Paco was driving his scooter west with an initial velocity of 4 m/s. He accelerates at 0.5 m/s2 for 30 seconds.

KonstantinChe [14]

Answer:

V = 19m/s

Explanation:

Given the following data;

Initial velocity, U = 4m/s

Acceleration, a = 0.5m/s²

Time, t = 30 seconds

To find the final velocity, we would use the first equation of motion;

V = U + at

Where;

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

V = 4 + 0.5*30

V = 4 + 15

V = 19m/s

Therefore, his final velocity is 19 meters per seconds.

6 0

3 years ago

Read 2 more answers

1. Choose the correct answer 1. Casparian strips are present in the of the root. a) cortex b) pith c) pericycle d) endodermis 2.

zloy xaker [14]

Answer:

1 casparian strips are present in the root of endodermis.

2 the endarch condition is the character fature of stem.

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3 years ago

You are riding your bike to the mall. You travel the first mile in 10 minutes. The last mile takes you 15 minutes. This is an ex

Vikki [24]

This is an example of slowing down because you encountered heavy traffic or just got tired. Also, the numbers show acceleration.

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3 years ago