W = ∫ (x from 0.1 to +oo) F dx
= ∫ (x from 0.1 to +oo) A e^(-kx) dx
= A/k x [ - e^(-kx) ](between 0.1 and +oo)
= A/k x [ 0 + e^(-k * 0.1) ]
<span>
= A/k x e^(-k/10) </span>
<span>Hydrocarbons are molecules that contain only carbon and hydrogen.</span>
Due to carbon's unique bonding patterns, hydrocarbons can have single, double, or triple bonds between the carbon atoms.
The names of hydrocarbons with single bonds end in "-ane," those
with double bonds end in "-ene," and those with triple bonds end in
"-yne".
The bonding of hydrocarbons allows them to form rings or chains.
Weather balloons are filled with only a small amount of helium because the __Volume__. of the balloon will increase as the air pressure decreases at higher altitudes.
a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J
b.) K 2 +GmM( r 11− r 21)=6.9×10 7 J
Applying Law of Energy conservation :
K 1+U 1
=K 2+U 2
⇒K 1− r 1GmM
=K 2− r 2 GmM
where M=5.0×10 23kg,r1
=> R=3.0×10 6m and m=10kg
(a) If K 1
=5.0×10 7J and r 2
=4.0×10 6 m, then the above equation leads to
K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J
(b) In this case, we require K 2
=0 and r2
=8.0×10 6m, and solve for K 1:K 1
=K 2 +GmM (r 11− r 21)=6.9×10 7 J
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