9.0, might be to late but someone in the future might need it.
Melting. Water freezes at 0 degrees celsius or 32 degrees fahrenheit and therefore water freezes at that temperature or ice melts back to its liquid state at those temperatures.
Answer:
v_f = 10.38 m / s
Explanation:
For this exercise we can use the relationship between work and kinetic energy
W = ΔK
note that the two quantities are scalars
Work is defined by the relation
W = F. Δx
the bold are vectors. The displacement is
Δx = r_f -r₀
Δx = (11.6 i - 2j) - (4.4 i + 5j)
Δx = (7.2 i - 7 j) m
W = (4 i - 9j). (7.2 i - 7 j)
remember that the dot product
i.i = j.j = 1
i.j = 0
W = 4 7.2 + 9 7
W = 91.8 J
the initial kinetic energy is
Ko = ½ m vo²
Ko = ½ 2.0 4.0²
Ko = 16 J
we substitute in the initial equation
W = K_f - K₀
K_f = W + K₀
½ m v_f² = W + K₀
v_f² = 2 / m (W + K₀)
v_f² = 2/2 (91.8 + 16)
v_f = √107.8
v_f = 10.38 m / s
When analyzing inelastic collisions, we need to consider the law of conservation of momentum, which states that the total momentum, p, of the closed system is a constant. In the case of inelastic collisions, the momentum of the combined mass after the collision is equal to the sum of the momentum of each of the initial masses.
p1+p2+...=pf
In our case we only have two masses, which makes our problem fairly simple. Lets plug in the formula for momentum; p=mv.
m1v1+m2v2=(m1+m2)vf
To find the velocity of the combined mass we simply rearrange the terms.
vf=m1v1+m2v2m1+m2
Plug in the values given in the problem.
vf=(3.0kg)(1.4m/s)+(2.0kg)(0m/s)03.0kg+2.0kg
vf=.84m/s
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.
