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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Explanation:
<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>
Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass
as


Substituting (2) into (1), we get

where
, the frictional force on
Set this aside for now and let's look at the forces on 
<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>
Let the x-axis be (+) up along the inclined plane. We can write the forces on
as


From (5), we can solve for <em>N</em> as

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

Substituting (7) into (4) we get

Collecting similar terms together, we get

or
![a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)](https://tex.z-dn.net/?f=a%20%3D%20%5Cleft%5B%20%5Cdfrac%7Bm_B%5Csin30%20-%20%5Cmu_km_A%7D%7B%28m_A%20%2B%20m_B%29%7D%20%5Cright%5Dg%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%288%29)
Putting in the numbers, we find that
. To find the tension <em>T</em>, put the value for the acceleration into (7) and we'll get
. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get 
Answer: condensation process is exothermic. The correct option is IV.
Explanation: Exothermic reaction is a chemical reaction in which heat is released to the surrounding environment.
condensation is defined as a process by which a medium changes from gaseous phase into liquid phase. Using water molecules to illustrate, to transform water molecules to gaseous phase it requires the heat of vaporization to be overcomed. This same hear of vaporization needs to be given off when changing back to liquid phase.
<h2>
Option 3, 216 m is the correct answer.</h2>
Explanation:
We have initial velocity, u = 15 m/s
Time, t = 12 seconds
Final velocity, v = 21 m/s
We have equation of motion v = u + at
Substituting
21 = 15 + a x 12
a = 0.5 m/s²
Now we have equation of motion v² = u² + 2as
21² = 15² + 2 x 0.5 x s
s = 216 m
Displacement = 216 m
Option 3, 216 m is the correct answer.