Answer:
=20 turns
Explanation:
The given case is a step down transformer as we need to reduce 120 V to 6 V.
number of turns on primary coil N_{P}= 400
current delivered by secondary coil I_{S}= 500 mA
output voltage = 6 V (rms)
we know that
putting values we get
to calculate number of turns in secondary
therefore, =20 turns
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
You can solve this using the equation k=mv^2. k is the kinetic energy, m is the mass, and is the velocity. k=30(3.4)^2, so the answer is 346.8 J.
Answer:
1.0×10³ N
Explanation:
μs is the static coefficient of friction. That's the friction that acts on a stationary (non-moving) object when being pushed or pulled.
μk is the kinetic coefficient of friction. That's the friction that acts on a moving object.
To budge the pig (while it's still stationary), we need to overcome the static friction.
F = N μs
For a non-moving object on level ground, the normal force N equals the weight.
F = mg μs
Given m = 130 kg and μs = 0.80:
F = (130 kg) (9.8 m/s²) (0.80)
F = 1019.2 N
Rounded to two significant figures, the force needed to budge the pig is 1.0×10³ N.
