Answer:
Work done = 35467.278 J
Explanation:
Given:
Height of the cone = 4m
radius (r) of the cone = 1.2m
Density of the cone = 600kg/m³
Acceleration due to gravity, g = 9.8 m/s²
Now,
The total mass of the cone (m) = Density of the cone × volume of the cone
Volume of the cone = 
thus,
volume of the cone = 
= 6.03 m³
therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg
The center of mass for the cone lies at the 
times the total height
thus,
center of mass lies at, h' = 
Now, the work gone (W) against gravity is given as:
W = mgh'
W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J
Answer:
4.64m/s
Explanation:
We can use the formula [ v = √2gh ] to solve for this problem. We know that g is constant acceleration (9.8), and h is height (1.1).
v = √2(9.8)(1.1)
v ≈ 4.64m/s
Best of Luck!
Answer:
b) lattice energy
Explanation:
A solution is said to have colligative property when the property depends on the solute present in the solution.
Colligative property depend upon on the solute particle or the ion concentration not on the identity of solute.
osmotic pressure, vapor pressure lowering , boiling point elevation and freezing point lowering all depend upon solute concentration so they will not have colligative property so, the answer remains option 'b' which is lattice energy.
I would make the ramp flatter. In doing so the ramp would have to be longer.
The speed of A and B immediately after collision is 5.28m/s
<u>Explanation:</u>
Mass of A is 6275kg
Speed of A is 6.5m/s
Mass of B is 5155kg
Speed of B is 3.8m/s
Track is frictionless.
A and B stick together.
speed of attached A and B = ?
mₐsₐ + mᵇsᵇ = (mₐ + mb) s
Therefore, The speed of A and B immediately after collision is 5.28m/s
