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3 years ago

7

Match each vocabulary word with the correct definition. 1. measure of how quickly velocity is changing 2. speed in a given direc

tion 3. force that resists moving one object against another 4. measure of the pull of gravity on an object 5. tendency of an object to resist a change in motion 6. size friction acceleration velocity magnitude inertia weight friction

2 answers:

IrinaK [193]3 years ago

8 0

1. measure of how quickly velocity is changing . . . acceleration

2. speed in a given direction . . . velocity

3. force that resists moving one object against another . . . friction

4. measure of the pull of gravity on an object . . . weight

5. tendency of an object to resist a change in motion . . . inertia

6. size . . . magnitude

Mars2501 [29]3 years ago

4 0

Answer:

1. measure of how quickly velocity is changing - acceleration

2. speed in a given direction - velocity

3. force that resists moving one object against another - friction

4. measure of the pull of gravity on an object - weight

5. tendency of an object to resist a change in motion - inertia

6. size - magnitude

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Explanation:

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

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$m_{e} = 9.1\times 10_{- 31} kg$ = 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):

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

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3 years ago

What is the pressure at a depth of 15 cm brine of density 1.2/cm³?

Anna [14]

P = density × gravity acceleration × height

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6 0

2 years ago

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netineya [11]

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Before the firecracker blows a coconut does not move, so left side is equal to 0:
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We know that m1=m2=m and m3=2m. Also we are asked to find v3f so we can rewrite formula:
$v_{3f} = - \frac{m_{1} v_{1f} + m_{2} v_{2f} }{ m_{3} }$

We must take in consideration that two parts with same mass do not move in same direction. The center of mass of these two parts moves between them at angle of 45° with respect to both south and west. The speed of a center of mass is:
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This speed we can insert into formula for v3f:
$v_{3f} = - \frac{m*33.9+m*33.9 }{ 2m } \\ \\ v_{3f} = - \frac{2m*33.9 }{ 2m } \\ \\ v_{3f} = - 33.9m/s$

We can see that part of a coconut with biggest mass has same speed as center of mass of two other parts. Negative sign shows that direction is opposite to direction of two pats. Biggest part moves towards north-east.

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3 years ago

Read 2 more answers