Answer:
I will choose the answer A
Looking at the force-time graph, wouldn't the force be integral fdt between 0 and 10s, a sort of "smoothed out pulse" ? And it looks like a familiar bell shaped curve. doesn't that produce a turning effect/torque and isn't there something about a circular analogue to F=ma in newtonian linear mechs ???
Looks like a difficult question for 5 points
Answer:
Explanation:
No external force is acting on uranium nucleus so we can apply law of conservation of momentum . If M and m be the mass of bigger and smaller fragments respectively and V and v be their velocity after fission , Total momentum after fission = total momentum before fission
MV - mv = 0 , since the nucleus was at rest , its momentum before fission is zero . negative sign of mv is taken because smaller fragment will move in opposite direction to that of bigger fragment .
MV = mv
M / m = v / V
M > m
Hence,
v > V
So velocity of smaller fragment will be higher .
Your question has been heard loud and clear.
An alpha particle , can move in any direction randomly. But with a magnetic field , we can deflect the alpha particle in any direction we want.
So , the magnetic field must be placed to the west of the alpha particle , so that the particle gets deflected and moves towards the north direction.
Thank you.
Answer:
a) ΔV = 2,118 10⁻⁸ m³ b) ΔR= 0.0143 cm
Explanation:
a) For this part we use the concept of density
ρ = m / V
As we are told that 1 carat is 0.2g we can make a rule of proportions (three) to find the weight of 2.8 carats
m = 2.8 Qt (0.2 g / 1 Qt) = 0.56 g = 0.56 10-3 kg
V = m / ρ
V = 0.56 / 3.52
V = 0.159 cm3
We use the relation of the bulk module
B = P / (Δv/V)
ΔV = V P / B
ΔV = 0.159 10⁻⁶ 58 10⁹ /4.43 10¹¹
ΔV = 2,118 10⁻⁸ m³
b) indicates that we approximate the diamond to a sphere
V = 4/3 π R³
For this part let's look for the initial radius
R₀ = ∛ ¾ V /π
R₀ = ∛ (¾ 0.159 /π)
R₀ = 0.3361 cm
Now we look for the final volume and with this the final radius
= V + ΔV
= 0.159 + 2.118 10⁻²
= 0.18018 cm3
= ∛ (¾ 0.18018 /π)
= 0.3504 cm
The radius increment is
ΔR = - R₀
ΔR = 0.3504 - 0.3361
ΔR= 0.0143 cm