You drive on Interstate 10 from San Antonio to Houston, half the time at 72 km/h and the other half at 98 km/h. On the way back
1 answer:
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Answer:
F_total = 29.4 N, directed to the right of particle 2
Explanation:
We must solve this problem in parts, first we calculate each force and then we apply Newton's law to add the forces.
Let's use Coulomb's law to calculate each force
F =
particles 1 and 2
q₁ = 8.0 10⁻⁶ C, q₂ = 3.5 10⁻⁶ C x₁₂ = 0.10 m
F₁₂ = 9 10⁹ 8.0 3.5 10⁻¹² / 0.1²
F₁₂ = 2.59 10¹ N
Since the two charges are of the same sign, this force is repulsive and is directed towards the positive side of the x axis.
particles 2 and 3
q₂ = 3.6 10⁻⁶ C, q₃ = 2.5 10⁻⁶ C, x₂₃ = 0.15 m
we calculate
F₂₃ = 9 10⁹ 3.5 2.5 10⁻¹²/ 0.15²
F₂₃ = 3.5 N
as the charge is of different sign, the force is attractive, therefore it is directed to the right of the load 2
Now we add the forces as vectors
F_total = ∑ F = F₁₂ + F₂₃
F_total = 25.2 +3.5
F_total = 29.4 N
directed to the right of particle 2
Answer:
3349J/kgC
Explanation:
Questions like these are properly handled having this fact in mind;
- Heat loss = Heat gained
Quantity of heat = mcΔ∅
m = mass of subatance
c = specific heat capacity
Δ∅ = change in temperature
m₁c₁(∅₂-∅₁) = m₂c₂(∅₁-∅₃)
m₁ = mass of block = 500g = 0.5kg
c₁ = specific heat capacity of unknown substance
∅₂ = block initial temperature = 50oC
∅₁ = equilibrium temperature of block and water after mix= 25oC
m₂= mass of water = 2kg
c₂ = specific heat capacity of water = 4186J/kg C
∅₃ = intial temperature of water = 20oC
0.5c₁(50-25) = 2 x 4186(25-20)
And we can find c₁ which is the unknown specific heat capacity
c₁ = = 3348.8J/kg C≅ 3349J/kg C