In series with the circuit, so for it pass the current to be mensured.
Letter A
If you notice any mistake in my english, please let me know, because i am not native.
B.strong but weaker than the weak force
Answer:
187 J
Explanation:
First Law of Thermodynamics :
ΔQ = ΔW + ΔU
ΔQ : Heat. If it added to system then positive and if it is rejected by system then negative.
ΔW : Work. If it done by the system then positive and if it is done on system then negative.
ΔU : Internal Energy. If it positive then temperature of system increased and if it is negative then temperature of system decreased.
ΔQ = 79 J
ΔW = - 108 J
ΔU = ?
substituting the value in the equation:
79 = -108 + ΔU
∴ ΔU = 187 J
Answer:
P₂ = 138.88 10³ Pa
Explanation:
This is a problem of fluid mechanics, we must use the continuity and Bernoulli equation
Let's start by looking for the top speed
Q = A₁ v₁ = A₂ v₂
We will use index 1 for the lower part and index 2 for the upper part, let's look for the speed in the upper part (v2)
v₂ = A₁ / A₂ v₁
They indicate that A₂ = ½ A₁ and give the speed at the bottom (v₁ = 1.20 m/s)
v₂ = 2 1.20
v₂ = 2.40 m / s
Now let's write the Bernoulli equation
P₁ + ½ ρ v₁² + ρ g y₁ = P2 + ½ ρ v₂² + ρ g y₂
Let's clear the pressure at point 2
P₂ = P₁ + ½ ρ (v₁² - v₂²) + ρ g (y₁-y₂)
we put our reference system at the lowest point
y₁ - y₂ = -20 cm
Let's calculate
P₂ = 143 10³ + ½ 1000 (1.20² - 2.40²) + 1000 9.8 (-0.200)
P₂ = 143 103 - 2,160 103 - 1,960 103
P₂ = 138.88 10³ Pa
