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2 years ago

Which of the following can effect LRO?

Engineering

1 answer:

aleksley [76]2 years ago

7 0

Answer:

The lunar radiation environment, allowing scientists to determine potential impacts to astronauts and other life. It also will test models on the effects of radiation and measure radiation absorption by a type of plastic that is like human tissue. The results could aid in the development of protective technologies to help keep future lunar crew members safe. CRaTER was built and developed by Boston University and the Massachusetts Institute of Technology in Boston.

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What is the command in the java script that receives data from the user

solmaris [256]

Answer:ummmmmmmmmmmm

Explanation:

Uuaysgggahanwnyecegvxg

6 0

3 years ago

The diameter of a cylindrical water tank is Do and its height is H. The tank is filled with water, which is open to the atmosphe

Sonbull [250]

Answer:

a. The time required for the tank to empty halfway is presented as follows;

$t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)$

b. The time it takes for the tank to empty the remaining half is presented as follows;

$t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }$

The total time 't', is presented as follows;

$t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }$

Explanation:

a. The diameter of the tank = D₀

The height of the tank = H

The diameter of the orifice at the bottom = D

The equation for the flow through an orifice is given as follows;

v = √(2·g·h)

Therefore, we have;

$\dfrac{P_1}{\gamma} + z_1 + \dfrac{v_1}{2 \cdot g} = \dfrac{P_2}{\gamma} + z_2 + \dfrac{v_2}{2 \cdot g}$

$\left( \dfrac{P_1}{\gamma} -\dfrac{P_2}{\gamma} \right) + (z_1 - z_2) + \dfrac{v_1}{2 \cdot g} = \dfrac{v_2}{2 \cdot g}$

Where;

P₁ = P₂ = The atmospheric pressure

z₁ - z₂ = dh (The height of eater in the tank)

A₁·v₁ = A₂·v₂

v₂ = (A₁/A₂)·v₁

A₁ = π·D₀²/4

A₂ = π·D²/4

A₁/A₂ = D₀²/(D²) = v₂/v₁

v₂ = (D₀²/(D²))·v₁ = √(2·g·h)

The time, 'dt', it takes for the water to drop by a level, dh, is given as follows;

dt = dh/v₁ = (v₂/v₁)/v₂·dh = (D₀²/(D²))/v₂·dh = (D₀²/(D²))/√(2·g·h)·dh

We have;

$dt = \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } dh$

The time for the tank to drop halfway is given as follows;

$\int\limits^{t_1}_0 {} \, dt = \int\limits^h_{\frac{h}{2} } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh$

$t_1 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{\frac{H}{2} }^{H} =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{\frac{H}{2} }^{H} = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)$

$t_1 = { \dfrac{2 \cdot D_0^2 }{D^2\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)$

$t_1 = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{2 \cdot H} - \sqrt{{H} } \right) =\dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)$ The time required for the tank to empty halfway, t₁, is given as follows;

$t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)$

(b) The time it takes for the tank to empty completely, t₂, is given as follows;

$\int\limits^{t_2}_0 {} \, dt = \int\limits^{\frac{h}{2} }_{0 } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh$

$t_2 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{0}^{\frac{H}{2} } =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{0 }^{\frac{H}{2} } = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left( \sqrt{\dfrac{H}{2} } -0\right)$

$t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }$

The time it takes for the tank to empty the remaining half, t₂, is presented as follows;

$t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }$

The total time, t, to empty the tank is given as follows;

$t = t_1 + t_2 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right) + t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} } = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \sqrt{2}$

$t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }$

3 0

2 years ago

प्रहार का समरूपी भिन्नार्थक शब्द अर्थ के साथ

Reika [66]

Answer:

uhvdiuuauivhsifuhviudshivuhrwwwwwgyuvrrrrrrvuybgdwdoushduhsfuhdsiufhduisfhsofhuaifhuadsihfuiahfuidhsfiuhdsiufhdsaigsdygfewyihfiewbvhwegfiyerrrgqewfygediyfgdsyifguysdgfyuuaegfuyaeggfuyegwfyugewafuyegwiufggeegfyusdggfsgodqwufyew98fyyy79ewyf7eywf8777etf87tewtf

Explanation:

djhidhfdiushfiudshfui

3 0

3 years ago

This question is 100 points<br> I NEED HELP!!!

Mamont248 [21]

Answer:

hey if u repost this i can answer it u and u dont have to waste this much points but its super blury and not even able to read a single word

8 0

3 years ago

Read 2 more answers

A silicon diode has a saturation current of 6 nA at 25 degrees Celcius. What is the saturation current at 100 degrees Celsius?