Physics
What is the momentum of a 1.5 × 103 kilogram van that is moving at a velocity of 32 meters/second? A. 46.9 kilogram meters/second B. 4.7 × 103 kilogram meters/second C. 4.85 × 102 kilogram meters/second D. 4.85 × 104 kilogram meters/second
Answer:
pu = 1260.9kg/m^3
the density of the unknown liquid is 1260.9kg/m^3
Explanation:
The density of a liquid is inversely proportional to the volume (height) of object submerged in it.
High density liquid possess higher buoyant force preventing objects from submerging.
p ∝ 1/V ∝ 1/h
since V = Ah
pu/pw = hw/hu
pu = pwhw/hu
Where;
p = density
h = height submerged
pu and pw is the density of unknown liquid and water respectively
hu and hw is the height of object submerged in unknown liquid and water respectively
pw = 1000kg/m^3
hu = 4.6cm = 0.046m
hw = 5.8cm = 0.058m
Substituting the given values;
pu = 1000×0.058/0.046
pu = 1260.9kg/m^3
the density of the unknown liquid is 1260.9kg/m^3
Answer:

Explanation:
Given that:
mass of object A, 
mass of object B, 
speed of object A, 
So, according to the conservation of momentum, the momentum before collision is equal to the momentum after conservation.




Answer:
C
Explanation:
To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):
d = d₀ + d₀αT
for the sphere, we were given
D₀ = 4.000 cm
α = 1.1 x 10⁻⁵/degrees celsius
we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T EQN 1
Similarly for the Aluminium ring we have
we were given
d₀ = 3.994 cm
α = 2.4 x 10⁻⁵/degrees celsius
we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T EQN 2
Since @ the temperature T at which the sphere fall through the ring, d=D
Eqn 1 = Eqn 2
4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms
0.006=5.18x10⁻⁵T
T=115.7K