A lever is used to lift a heavy weight. The length of the input arm of the lever is 3 m, the length of output arm is 1 m.

Two identical strings, of identical lengths of 2.00 m and linear mass density of μ=0.0065kg/m, are fixed on both ends. String A

kolezko [41]

Answer:

beat frequency = 13.87 Hz

Explanation:

given data

lengths l = 2.00 m

linear mass density μ = 0.0065 kg/m

String A is under a tension T1 = 120.00 N

String B is under a tension T2 = 130.00 N

n = 10 mode

to find out

beat frequency

solution

we know here that length L is

L = n × $\frac{ \lambda }{2}$ ........1

so λ = $\frac{2L}{10}$

and velocity is express as

V = $\sqrt{\frac{T}{\mu } }$ .................2

so

frequency for string A = f1 = $\frac{V1}{\lambda}$

f1 = $\frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}$

f1 = $\frac{10}{2L} \sqrt{\frac{T1}{\mu } }$

and

f2 = $\frac{10}{2L} \sqrt{\frac{T2}{\mu } }$

so

beat frequency is = f2 - f1

put here value

beat frequency = $\frac{10}{2*2} \sqrt{\frac{130}{0.0065}}$ - $\frac{10}{2*2} \sqrt{\frac{120}{0.0065} }$

beat frequency = 13.87 Hz

6 0

4 years ago

The planets eccentricity in order least to greatest

ankoles [38]

Answer:

Orbital Eccentricity

Planet Orbital Eccentricity

(Point in Orbit Closest to Sun)

measured in AU's

Mercury 0.206

Venus 0.007

Earth 0.017

Mars 0.093

Jupiter 0.048

Saturn 0.056

Uranus 0.047

Neptune 0.009

Pluto 0.248

Explanation:

link to information:

https://www.enchantedlearning.com/subjects/astronomy/glossary/Eccentricity.shtml

3 0

2 years ago

An athlete at high performance inhales 4.0L of air at 1 atm and 298 K. The inhaled and exhaled air contain 0.5% and 6.2% by volu

LenaWriter [7]

To solve this problem we will calculate the total volume of inhaled and exhaled water. From the ideal gas equation we will find the total number of moles of water.

An athlete at high performance inhales 4.0L of air at 1atm and 298K.

The inhaled and exhaled air contain 0.5% and 6.2% by volume of water, respectively.

During inhalation, volume of water taken is

$V_i = (4L)(0.5\%)$

$V_i = 0.02L$

During exhalation, volume of water expelled is

$V_e = (4L)(6.2\%)$

$V_e = 0.248L$

During 40 breathes, total volume of water taken is

$V_{it} = (40L)(0.02L) = 0.8L$

During 40 breathes, total volume of water expelled out is

$V_{et} = (40L)(0.248L) = 9.92L$

Therefore resultant volume of water expelled out from the lung is

$\Delta V = 9.92L-0.8L = 9.12$

From the body through the lung we have that

$n = \frac{PV}{RT}$

Here,

P = Pressure

R= Gas ideal constant

T= Temperature

V = Volume

Replacing,

$n = \frac{(1atm)(9.12L)}{(8.314J/mol \cdot K)(298K)}$

$n = 0.373mol/min$

Therefore the moles of water per minute are expelled from the body through the lungs is 0.373mol/min

8 0

4 years ago

If the frequency of this beam is increased while the intensity is held constant, does the number of electrons ejected per second

Ivan

Answer:

if the intensity of photons is constant then number of ejected electrons will remain same

Explanation:

As per photoelectric effect we know that when light of sufficient frequency fall on the surface of metal then electrons get ejected out of the surface with certain kinetic energy

Here the energy of photons is used to eject out the electrons from metal surface and to give the kinetic energy to the ejected electrons

so we have

$h\nu = W + KE$

here W = work function of metal which shows the energy required to eject out electrons from metal surface

KE = kinetic energy of ejected electrons

now if we increase the frequency of the photons that incident on the metal surface then in that case the incident energy will increase

So the electrons will eject out with more kinetic energy while if the number of photon is constant or the intensity of photons is constant then number of ejected electrons will remain same

8 0

3 years ago

Which pm the following are likely to form a covalent bond

Alecsey [184]

Answer:

magneisuim and gold

Explanation:

8 0

4 years ago

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