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serious [3.7K]
3 years ago
11

If velocity is 0.2m/s, what is the kinetic energy?

Physics
1 answer:
Natalija [7]3 years ago
3 0

\huge\boxed{♧ \: \mathfrak{ \underline{Answer} \: ♧}}

we know,

\boxed{kinetic  \:  \: energy =  \frac{1}{2} m {v}^{2} }

So,

\longmapsto \dfrac{1}{2}  \times 10 \times 0.2 \times 0.2

\longmapsto0.2 \: joules

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Why do adults make bigger splashes when they jump into swimming pools than small children?
Semenov [28]

Answer:

it bc the adults are bigger then us kid so when they dip in the pool it makes bigger splashes

Explanation:

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3 years ago
What type of plate boundary decreases the amount of the Earth's crust?
rosijanka [135]

Answer:

Convergent.

Explanation:

Just as oceanic crust is formed at mid-ocean ridges, it is destroyed in subduction zones. Subduction is the important geologic process in which a tectonic plate made of dense lithospheric material melts or falls below a plate made of less-dense lithosphere at a convergent plate boundary.

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2 years ago
What are all the<br> invertebrates with a<br> large foot
lyudmila [28]

Answer:

Explanation:

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3 years ago
Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3
cupoosta [38]

Answer:

844°C

Explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

J = - D \frac{\Delta c}{\Delta x}

Where \frac{\Delta c}{\Delta x} is the concentration of gradient

D is the diffusivity coefficient

and J is the flux of atoms.

In the other hand we have, that

D= D_0 e^{\frac{E_d}{RT}}

Where D_0 is the proportionality constant,

R is the gas constant, T the temperature and E_d is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}

Rearrange the equation to get the value of temperature,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

We have all the values in our equation.

\Delta c = 0.664-0.339 = 0.325 C. cm^{-1}

\Delta x = 9.7*10^{-3}m

E_d = 82000J

D_0 = 6.5*10^{-7}m^2/s

J = 3.2*10^{-9}m^2/s

R= 8.31Jmol^{-1}K

Substituting,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

T=-\frac{-82000}{(8.31)ln(\frac{3.2*10^{-9}(9.7*10^{-3})}{6.5*10^{-7} (0.325)})}

T=1118.07K=844\°C

4 0
3 years ago
In 'coin on card' experiment a smooth card is used. ​
KIM [24]

Answer:

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3 years ago
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