Why does oil not dissolve in water?
1 answer:
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The problem is solved and the questions are answered below.
Explanation:
a. To calculate the speed of the 0.66 kg ball just before the collision
V₀ + K₀ = V₁ + K₁
= mgh₀ = 1/2 mv₁²
where, h= r - r cosθ
V =
V = 2.42 m/s
b. Calculate the speed of the 0.22 kg ball immediately after the collision
y = y₀ + Vy₀t - 1/2 gt²
0 = 1.2 - 1/2 gt²
t = 0.495 s
x = x₀ + Vx₀t
1.4 = 0 + vx₀ (0.495)
Vx₀ = 2.83 m/s
C. To Calculate the speed of the 0.66 kg ball immediately after the collision
m₁ v₁ = m₁ v₃ + m₂ v₄
(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)
V₃ = 1.48 m/s
D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.
E. To Calculate the height to which the 0.66 kg ball rises after the collision
V₀ + k₀ = V₁ + k₁
1/2 mv₀² = mgh₁
h₁ = v₀²/2 g
= 0.112 m
F. Based on your data, No the collision is not elastic.
Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²
= 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²
= - 0.329 J
Hence, kinetic energy is not conserved.
Answer:
(a) 43.2 kC
(b) 0.012V kWh
(c) 0.108V cents
Explanation:
<u>Given:</u>
- i = current flow = 3 A
- t = time interval for which the current flow =
- V = terminal voltage of the battery
- R = rate of energy = 9 cents/kWh
<u>Assume:</u>
- Q = charge transported as a result of charging
- E = energy expended
- C = cost of charging
Part (a):
We know that the charge flow rate is the electric current flow through a wire.
Hence, 43.2 kC of charge is transported as a result of charging.
Part (b):
We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:
Hence, 0.012V kWh is expended in charging the battery.
Part (c):
We know that the energy cost is equal to the product of energy expended and the rate of energy.
Hence, 0.108V cents is the charging cost of the battery.