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Ivanshal [37]
2 years ago
9

Why do you think it is helpful for the widest part of a leaf to face the sky

Physics
1 answer:
VladimirAG [237]2 years ago
4 0
Become president of because leaf
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What is Initial temperature and final temperature equations??<br> ...?
Neporo4naja [7]
One that can help you is:
ΔT=<span>T<span>Final</span></span>−<span>T<span>Initia<span>l
That is of course adding both tmepratures. There is one more that is a lil bit more complex 
</span></span></span><span><span>Tf</span>=<span>Ti</span>−Δ<span>H<span>rxn</span></span>∗<span>n<span>rxn</span></span>/(<span>C<span>p,water</span></span>∗<span>m<span>water</span></span>)
This one is taking into account that yu can find temperature and that there could be a change with a chemical reaction. Hope this helps</span>
8 0
3 years ago
What is the period of a 4.12 m long pendulum?
lisov135 [29]
Using the equation for period length, you get an answer of about 4.1 seconds.
8 0
3 years ago
How much energy (in kilojoules) is required to convert 200 mL of diethyl ether at its boiling point from liquid to vapor if its
LiRa [457]

Answer:

55.96kJ

Explanation:

Energy = mass of diethyl ether × enthalpy of vaporization of diethyl ether

Volume (v) = 200mL, density (d) = 0.7138g/mL

Mass = d × v = 0.7138 × 200 = 142.76g

Enthalpy of vaporization of diethyl ether = 29kJ/mol

MW of diethyl ether (C2H5)2O = 74g/mol

Enthalpy in kJ/g = 29kJ/mol ÷ 74g/mol = 0.392kJ/g

Energy = 142.76g × 0.392kJ/g = 55.96kJ

4 0
3 years ago
Please help me with the equations for this! Three uniform spheres are fixed at the positions shown in the diagram. ( there is a
lora16 [44]
The change in gravitational potential energy due to change in position must be the change in it's kinetic energy as the system is isolated! so find out the potential energies of the two different points!

<span>PE=−[G<span>M1</span><span>M2</span>]÷R

</span><span> Potential energy of a particle due to mass A is not affected by presence of any other mass B !</span>
7 0
3 years ago
In the year 1178. five monks at Canterbury Cathedral in England observed what appeared to be an asteroid colliding with the moon
Lubov Fominskaja [6]

Answer: 1.28 sec

Explanation:

Assuming that the glow following the collision was produced instantaneously, as the light propagates in a straight line from Moon to the Earth at a constant speed, we can get the time traveled by the light applying velocity definition as follows:

V = ∆x / ∆t

Solving for ∆t, we have:

∆t = ∆x/v = ∆x/c = 3.84 108 m / 3.8 108 m/s = 1.28 sec

8 0
2 years ago
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