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Natasha_Volkova [10]

3 years ago

8

Bodies weighing 1 kilogram and 5 kilograms lie on a smooth horizontal surface. If a traction force of 0.6 N acts on another 5 kg

body, determine the acceleration of these bodies.

1 answer:

natima [27]3 years ago

6 0

0.6/5,1,5

so calculate it

not so sure though

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I stretch a rubber band and "plunk" it to make it vibrate in its fundamental frequency. I then stretch it to twice its length an

Nikitich [7]

Answer:

The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)

Explanation:

Fundamental frequency = wave velocity/2L

where;

L is the length of the stretched rubber

Wave velocity = $\sqrt{\frac{T}{\frac{M}{L}}}$

Frequency (F₁) = $\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}$

To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.

Given:

L₂ =2L₁ = 2L

T₂ = 2T₁ = 2T

(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)

F₂ = $\frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1$

Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).

7 0

3 years ago

if you knew the number of valence electrons in a nonmetal atom, how would you determine the valence of the element?

Vanyuwa [196]

It would be funny because . I will not be good

8 0

3 years ago

14. The average speed of a car was 60 m/s by the time it reached the finish line. The car moved in a straight line and traveled

Leni [432]

7.5m. Have a good day :)

4 0

2 years ago

A train whose proper length is 1200 m passes at a high speed through a station whose platform measures 900 m, and the station ma

TiliK225 [7]

Answer:

0.66c

Explanation:

Use length contraction equation:

L = L₀ √(1 − (v²/c²))

where L is the contracted length,

L₀ is the length at 0 velocity,

v is the velocity,

and c is the speed of light.

900 = 1200 √(1 − (v²/c²))

3/4 = √(1 − (v²/c²))

9/16 = 1 − (v²/c²)

v²/c² = 7/16

v = ¼√7 c

v ≈ 0.66 c

6 0

3 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

7 0

3 years ago