Refer to the figure below.
R = resistance.
Case 1:
The voltage source is V₁ and the current is 10 mA. Therefore
V₁ = (10 mA)R
Case 2:
The voltage source is V₂ and the current is 8 mA. Therefore
V₂ = (8 mA)R
Case 3:
The voltage across the resistance is V₁ - V₂. Therefore the current I is given by
V₁ - V₂ = IR
10R - 8R = (I mA)R
2 = I
The current is 2 mA.
Answer: 2 mA
Answer:
a) F = 2.66 10⁴ N, b) h = 1.55 m
Explanation:
For this fluid exercise we use that the pressure at the tap point is
Exterior
P₂ = P₀ = 1.01 105 Pa
inside
P₁ = P₀ + ρ g h
the liquid is water with a density of ρ=1000 km / m³
P₁ = 0.85 1.01 10⁵ + 1000 9.8 5
P₁ = 85850 + 49000
P₁ = 1.3485 10⁵ Pa
the net force is
ΔP = P₁- P₂
Δp = 1.3485 10⁵ - 1.01 10⁵
ΔP = 3.385 10⁴ Pa
Let's use the definition of pressure
P = Fe / A
F = P A
the area of a circle is
A = pi r² = [i d ^ 2/4
let's reduce the units to the SI system
d = 100 cm (1 m / 100 cm) = 1 m
F = 3.385 104 pi / 4 (1) ²
F = 2.66 10⁴ N
b) the height for which the pressures are in equilibrium is
P₁ = P₂
0.85 P₀ + ρ g h = P₀
h = 
h = 
h = 1.55 m
Answer:
It would be A, because it is has more height in which the potential energy would be greater.
Answer:
b. Static > sliding > rolling friction.
Explanation:
Static friction is greater than sliding friction. It takes more force to get an object to start sliding than to keep it sliding.
Sliding friction is greater than rolling friction. There are fewer points of contact for a round surface compared to a flat one.
Answer:
v = 3.27 m/s
Explanation:
KE = 1/2 mv^2
695 J = 1/2 (130kg)(v^2)
695 J / (1/2 x 130kg) = v^2
v^2 = square root of 10.69
v = 3.27 m/s
