| 1 |
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D |
|
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|
7 |
-.50
-.25
+.25
เต็ม
0
-35%
+30%
+35%
|
| 2 |
|
b |
|
|
|
7 |
-.50
-.25
+.25
เต็ม
0
-35%
+30%
+35%
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| 3 |
|
C |
|
|
|
7 |
-.50
-.25
+.25
เต็ม
0
-35%
+30%
+35%
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| 4 |
|
A |
|
|
|
7 |
-.50
-.25
+.25
เต็ม
0
-35%
+30%
+35%
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| 5 |
|
D |
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When the particles hit each other, the energy will be transferred from the moving particle to the particle at rest and the particle at rest will start moving.
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Conservation Of Momentum
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7 |
-.50
-.25
+.25
เต็ม
0
-35%
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| 6 |
|
E |
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The astronaut’s throw has a negligible impact on the total energy of the trash, and hence a negligible impact on a and hence T. After one full orbit, all 3 pieces of trash return to space station.
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Kepler’s 3rd Law, T^2 α a^3 where a is the semimajor axis, E = -GMm/2a.
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7 |
-.50
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-35%
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| 7 |
|
E |
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A and B is stationary. Center has 0 speed, then the highest point Q must have the highest speed.
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Vtop=v (2+ √3) / √3
|
7 |
-.50
-.25
+.25
เต็ม
0
-35%
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| 8 |
|
E |
|
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E0 = might +1/2 mv0^2 = 2Ef= 2mg(2h). So initial velocity we get v0 = √6th
|
7 |
-.50
-.25
+.25
เต็ม
0
-35%
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| 9 |
|
D |
|
|
|
7 |
-.50
-.25
+.25
เต็ม
0
-35%
+30%
+35%
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| 10 |
|
B |
|
|
|
7 |
-.50
-.25
+.25
เต็ม
0
-35%
+30%
+35%
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| 11 |
|
D |
|
|
|
7 |
-.50
-.25
+.25
เต็ม
0
-35%
+30%
+35%
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| 12 |
|
A |
|
|
|
7 |
-.50
-.25
+.25
เต็ม
0
-35%
+30%
+35%
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| 13 |
|
B |
|
|
|
7 |
-.50
-.25
+.25
เต็ม
0
-35%
+30%
+35%
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| 14 |
|
C |
|
|
|
7 |
-.50
-.25
+.25
เต็ม
0
-35%
+30%
+35%
|
| 15 |
|
C |
|
|
(δk/k)^2 = (δF/F)^2 + (δx /Δx)^2
|
7 |
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| 16 |
|
A |
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ΔP = 0 regardless of reference frame, ΔP doesn’t depend on frame. In an elastic collision, ΔK = 0 which also doesn’t depend on frame.
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ΔP = 0, ΔK=0
|
7 |
-.50
-.25
+.25
เต็ม
0
-35%
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| 17 |
|
E |
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Rotational angular velocity is ω. Bottom of cylinder is moving at 2ωr. The first plate is moving to the right at 2ωr.The second plate moves to the right 2ωr faster than the first plate and the top plate moves to the right at 6ωr
|
|
7 |
-.50
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เต็ม
0
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| 18 |
|
B |
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Force has 3 components 1. Shrek’s weight 2. Weight of water collected in ogre hat and 3. Pressure associated with momentum of downward falling rain. 1. Gives constant weight 5000N 2. Is equal to the weight of water collected in the hat.Volume of water: (1mm/s)*a*t & weight of water: V*(1000kg/m3)*g=10*t . 3. can be found by M = ρV of water that hits the hat has its momentum changed from p =Mv to 0. Here ρ = 1000 kg/m3 is the density of water. The force applied by the hat on the water is therefore F =Δp/Δt = Mv/t = 1N. Add all together gives total weight: 5000+1+10t = 5001+10t
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|
7 |
-.50
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+.25
เต็ม
0
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| 19 |
|
D |
|
|
|
7 |
-.50
-.25
+.25
เต็ม
0
-35%
+30%
+35%
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| 20 |
|
A |
|
gauge pressure is 0 at the top, so it is p1h1g at the liquid interface. Pressure then increases from the bottom giving the answer
|
p1h1g then increase
|
7 |
-.50
-.25
+.25
เต็ม
0
-35%
+30%
+35%
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