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2 months ago

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Ability of the muscles to function effectively and efficiently without undue fatigue

1 answer:

Keith_Richards [3.2K]2 months ago

7 0

Response:

Physical well-being

Clarification:

You might be interested in

In a demonstration, a 4.00 cm2 square coil with 10 000 turns enters a larger square region with a uniform 1.50 T magnetic field

serg [3582]

Explanation:

It is stated that,

Area of square coil, $A=4\ cm^2=0.0004\ m^2$

Length of one side of the square, L = 0.02 m

Number of coils, N = 10000

Consistent magnetic field, B = 1.5 T

Velocity, v = 100 m/s

An electromotive force is produced in the coil calculated as:

$E=NBLv$

$E=10000\times 1.5\times 0.02\times 100$

E = 30000 V

Breakdown voltage of air, $V=4000\ V/cm=400000\ V/m$

Let d be the distance between the ends of the coil wires that can still produce a spark. Therefore,

Electric field, $E'=\dfrac{V}{d}$

$\dfrac{30000}{d}=400000$

d = 0.075 m

Thus, this forms the final answer.

6 0

18 days ago

You are camping in the wilderness. After a few days, you are horrified to discover that you did not pack as many batteries as yo

Keith_Richards [3271]

Angular speed is calculated as 2.5 rev/sec.

8 0

28 days ago

A spin bike has a flywheel in two parts—a 12.5 kg disk with radius 0.23 m, and a 7.0 kg ring with mass concentrated at the outer

Keith_Richards [3271]

Response:

Clarification:

Provided

weight of disk $m=12.5 kg$

diameter of disc $R=0.23 m$

weight of ring $m_r=7 kg$

Force $F=9.7 N$

$N=180 rpm$

$\omega =\frac{2\pi N}{60}$

$\omega =6\pi rad/s$

Overall moment of inertia

=Disc's moment of inertia +Ring's Moment of Inertia

$=0.5\cdot 12.5\times 0.23^2+7\times 0.23^2$

$=13.25\times 0.23^2=0.7009 kg-m^2$

At this point, Torque is $T=F\times R=I\cdot \alpha$

$9.7\times 0.23=0.7\times \alpha$

$\alpha =3.18 rad/s^2$

Utilizing $\omega _f=\omega +\alpha t$

$\omega _f=0$ in this scenario

$0=6\pi -3.18\times t$

$t=\frac{6\pi }{3.18}$

$t=5.92 s$

7 0

25 days ago

3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag

serg [3582]

Response:

(a) $W=-19.25J$

(b) $W=-52.8J$

Clarification:

Greetings.

(a) In this case, since the starting volume is 18.5 dm³ and the ending volume is 21 dm³ (18.5 +2.5), we can calculate the work at constant pressure as shown below:

$W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J$

This value is negative as it expands against the given pressure.

(b) Furthermore, if the process is conducted reversibly, the pressure might change, hence, we need to calculate the work using:

$W=nRTln(\frac{V_1}{V_2} )$

The moles are calculated based on the provided mass of argon:

$n=6.56g*\frac{1mol}{39.95g}=0.164mol$

Consequently, the work amounts to:

$W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J$

Best regards.

4 0

16 days ago

Two of the types of ultraviolet light, uva and uvb, are both components of sunlight. their wavelengths range from 320 to 400 nm

Sav [3153]

In terms of light energy, a higher frequency corresponds to increased energy within the light.

We establish that frequency is essentially the inverse of wavelength:

frequency = 1 / wavelength

Calculating frequencies:

f UVA = 1/320 to 1/400

f UVA = 0.0031 to 0.0025

f UVB = 1/290 to 1/320

f UVB = 0.0034 to 0.0031

Since UVB occupies a higher frequency range, it consequently possesses greater energy than UVA.

7 0

1 month ago

Read 2 more answers