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Alexeev081 [22]
3 years ago
11

The process by which a characteristic spreads across space from one place to another over time is diffusion. True / False.

Physics
1 answer:
Fiesta28 [93]3 years ago
5 0

Answer:

True

Explanation:

True

Diffusion refers to that process through which concentration is transmitted from one place to another concentration i.e from higher concentration to lower concentration. Due to this only, characteristics able to spread all over the space.

it can only happen in fluid(liquid and gas) because their particle can move in a random direction

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In which example would the most transfer of energy take place from Mechanical energy to Thermal (heat) energy?
drek231 [11]

Answer:The answer is when a metal plate is pressed against a fast spinning piece of steel hard enough to stop it. i know this because I just took a test and got this question and got it write

Explanation:

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2 years ago
A 900-kg giraffe runs across a field at a rate of 50 km/hr. What is the magnitude of its momentum? 18 km/hr
Yuri [45]
The correct answer is the one with 45000 kg*km/hr.
the formula is p = m*v 
 900 *50/hr  giving u 45000 
I know this is the correct answer because i have already turned it in and got a 100%.  
7 0
3 years ago
Read 2 more answers
Two tuning forks produce sounds of wavelengths of 3.4 meters and 3.3 meters.
KengaRu [80]

Answer:

Explanation:

wavelength, λ = 3.4 m

wavelength, λ' = 3.3 m

Speed, v = 340 m/s

f = v / λ = 340 / 3.4 = 100 Hz

f' = v / λ' = 340 / 3.3 = 103.03 Hz

Frequency of beat, n = f' - f = 103.03 - 100 = 3.03 Hz

5 0
3 years ago
A wave with a low frequency generally has a _____.
Artyom0805 [142]
Its C because if it is a low frequency it will not change much so it will be a longer wavelength and the higher the frequency the shorter the wavelength
8 0
3 years ago
What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa
dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

E = \dfrac{hc}{\lambda}

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}

E=3.315\times10^{-17}\ J

(b).  We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

\lambda=\dfrac{h}{\sqrt{2mE}}

E=\dfrac{h^2}{2m\lambda^2}

Put the value into the formula

E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}

E=6.709\times10^{-21}\ J

(c). We need to calculate the energy of photon

Using formula of energy

E = \dfrac{hc}{\lambda}

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}

E=3.315\times10^{-11}\ J

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

\lambda=\dfrac{h}{\sqrt{2mE}}

E=\dfrac{h^2}{2m\lambda^2}

Put the value into the formula

E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}

E=6.709\times10^{-9}\ J

Hence, This is the required solution.

6 0
3 years ago
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