Answer:
This is an inelastic collision. This means, unfortunately, that KE cannot save you, at least in the problem's current form.
Let's see what conservation of momentum in both directions does ya:
Conservation in the x direction:
Only 1 object here has a momentum in the x direction initally.
m1v1i + 0 = (m1 + m2)(vx)
3.09(5.10) = (3.09 + 2.52)Vx
Vx = 2.81 m/s
Explanation:
Conservation in the y direction:
Again, only 1 object here has initial velocity in the y:
0 + m2v2i = (m1 +m2)Vy
(2.52)(-3.36) = (2.52 + 3.09)Vy
Vy = -1.51 m/s
++++++++++++++++++++
Now that you have Vx and Vy of the composite object, you can find the final velocity by doing Vf = √Vx^2 + Vy^2)
Vf = √(2.81)^2 + (-1.51)^2
Vf = 3.19 m/s
<span>When two hydrogen atoms bond, the positive nucleus of one atom attracts the electrons of both the atoms since the electrons are shared between the two atoms now.</span>
Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
Answer:
we need the graph to answer the question.
