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adell [148]
4 years ago
10

A unstretched spring hangs from the ceiling with a length of 0.31 m. a 9.1-kg block is hung from the spring and the spring stret

ches to be 0.50 m long. how long will the spring be if a 3.3-kg block is hung from the spring? for both cases, all vibrations of the spring are allowed to settle down before any measurements are made.
Physics
1 answer:
adoni [48]4 years ago
7 0

As we know that spring force is given by

F = kx

here we know that for the equilibrium of the mass suspended with the spring we have

mg = kx

so now we have

9.1(9.81) = k(0.50 -0.31)

now we have

k = 469.8 N/m

now we have another mass

m = 3.3 kg

again from above equation we have

3.3 (9.81) = 469.8(L - 0.31)

L = 0.38 m

so final length of the spring will be 0.38 m

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gtnhenbr [62]

Answer:

The correct answer might be r = 2.8^{27} meters.

Explanation:

<u>The Answer Given might not be correct, I just did what my brain said.</u>

As the angle is perpendicular so Θ=90.

Putting this in the equation to calculate the magnetic force as:

F = evBsinΘ

F= evBsin90                   *sin90 = 1 so,

F= evB.

Now when the electron will start to move in  a circle, The necessary force that makes the electron rotate in a circle is given by Centripetal force.

So,

    Magneteic Force = Centripetal Force

    evB = \frac{mv^{2} }{r}

    r = \frac{mv}{Be} ......(1)

Now the problem is, We don't know " v " so we need to calculate velocity first,

Calculation of Velocity:

                                   In order to calculate the velocity of electron, We should know the potiential difference with which the electrons are accelerated which in our case is 500ev. If "V" is the potiential difference, the energy gained by electrons during accelreation will be Ve. This appear as kinectic enrgy of electrons as,

         

                        K.E = Ve

                        \frac{1}{2}mv^{2} = Ve

                        v =  \sqrt{\frac{2ve}{m} }................(2)

Putting value of velocity in equation 2 from 1:

r = \frac{mv}{Be}  \sqrt{\frac{2ve}{m} }

r = \sqrt{\frac{2mev}{Be}}

r = \sqrt{\frac{(2)(9.1^{-31})(500) (1.6^{-19} )  }{ (1.75^{11} ) (10^{-4} ) } }

r = 2.8^{27} meters.

                               

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Answer:

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Explanation:

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The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu
Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

E=\phi + K

where

E=\frac{hc}{\lambda} is the energy of the incident light, with h being the Planck constant, c being the speed of light, and \lambda being the wavelength

\phi is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

\lambda=190 nm=1.9\cdot 10^{-7}m, so the energy of the incident light is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J

Converting in electronvolts,

E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

\phi = E-K=6.5 eV-4.0 eV=2.5 eV

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

\lambda=510 nm=5.1\cdot 10^{-7} m

So its energy is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J

Converting in electronvolts,

E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: \phi = 4.5 eV

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

E=\phi+K=4.5 eV+2.7 eV=7.2 eV

Then the copper is replaced with sodium, which has work function of

\phi = 2.3 eV

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

K=E-\phi = 7.2 eV-2.3 eV=4.9 eV

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Lady_Fox [76]
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Hope it helps!
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3 years ago
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