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2 years ago

Given this graph plotting velocity versus time, estimate the acceleration of object A at points X and Y respectively.

Physics

1 answer:

Nikolay [14]2 years ago

5 0

I think answer should be d. Please give me brainlest let me know if it’s correct

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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

only a relatively small part of the electromagnetic spectrum is visible. what determines which bands of the electromagnetic spec

swat32

Answer:

hsnzbssj

Explanation:

<h2>skibbbbbbbbbbbbibi</h2>

3 0

3 years ago

GUYS PLEASE HELP!! The number of individuals that die in a population in a certain time period is the _______ WHAT IS IT!? Pleas

ryzh [129]

I think it "Death rate" but I am not very sure though.

3 0

3 years ago

Which one of the following types of electromagnetic wave travels through space the fastest?

Setler [38]

Answer:

As a result, light travels fastest in empty space, and travels slowest in solids. In glass, for example, light travels about 197,000 km/s.

Explanation:

4 0

1 year ago

Why is the cycle of cellular respiration and photosynthesis important to life for plants

devlian [24]

The cycle of cellular respiration and photosynthesis are important for the life of plants because cellular respiration uses one of the products of photosynthesis (oxygen), and uses it as one of the reactants along with glucose to produce carbon dioxide. These two cycles help maintain a balance for our atmosphere, even though pollution and other factors have disrupted this balance. Hope this helps! :)

5 0

2 years ago