A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

The element carbon has 3 naturally occurring isotopes. About 99% of carbon isotopes are C-12, about 1% are C-13, and a tiny amou

mezya [45]

Average atomic mass of carbon is 12.01 amu

3 0

3 years ago

Ask Your Teacher In the air over a particular region at an altitude of 500 m above the ground, the electric field is 120 N/C dir

strojnjashka [21]

Answer:

$1.475\times 10^{-13}\ C/m^3$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

A = Area

h = Altitude = 600 m

Electric flux through the top would be

$-110A$ (negative as the electric field is going into the volume)

At the bottom

$120A$

Total flux through the volume

$\phi=120-110\\\Rightarrow \phi=10A$

Electric flux is given by

$\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=10A\epsilon_0$

Charge per volume is given by

$\rho=\dfrac{q}{v}\\\Rightarrow \rho=\dfrac{10A\epsilon_0}{Ah}\\\Rightarrow \rho=dfrac{10\epsilon_0}{h}\\\Rightarrow \rho=\dfrac{10\times 8.85\times 10^{-12}}{600}\\\Rightarrow \rho=1.475\times 10^{-13}\ C/m^3$

The volume charge density is $1.475\times 10^{-13}\ C/m^3$

7 0

3 years ago

Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const

kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is L= 20cm = 0.2m

Thermal conductivity of the wall is K = 2.79 W/m·K

Temperature at the left side surface is T₁ = 50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

$Q_{conduction} = Q_{convection}$

Expression for the heat conduction process is

$Q_{conduction} = \frac{K(T_1 - T)}{L}$

Expression for the heat convection process is

$Q_{convection} = h(T_2 - T)$

Substitute the expressions of conduction and convection in equation above

$Q_{conduction} = Q_{convection}$

$\frac{K(T_1 - T_2)}{L} = h(T_2 - T)$

Substitute the values in above equation

$\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC$

Now heat flux through the wall can be calculated as

$q_{flux} = Q_{conduction} \\\\q_{flux} = \frac{K(T_1 - T_2)}{L}\\\\q_{flux} = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2$

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

6 0

3 years ago

IF THERE ARE ONLY 118 ELEMENTS, HOW DO YOU ACCOUNT FOR THE MANY MILLIONS OF THINGS THAT WE HAVE IN OUR UNIVERSE?

bogdanovich [222]

Answer:

theres only 118 elements that are discovered. now that they're the only ones out there

Explanation:

3 0

3 years ago

What is a conclusion?

Vikki [24]

The end or finish of an event or process.

8 0

3 years ago

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