Answer:
a) a= 8.33 m/s², T = 12.495 N
, b) a = 2.45 m / s²
Explanation:
a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.
We apply Newton's second law to the lower charge
fr₁ + fr₂ = ma
The equation for the force of friction is
fr = μ N
Y Axis
N - W₁ –W₂ = 0
N = W₁ + W₂
N = (m₁ + m₂) g
Since the beams are the same, it has the same mass
N = 2 m g
We replace
μ₁ 2mg + μ₂ mg = m a
a = (2μ₁ + μ₂) g
a = (2 0.30 + 0.25) 9.8
a= 8.33 m/s²
Let's look for cable tension with beam 2
T = m₂ a
T = 1500 8.33
T = 12.495 N
b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal
fr = m₂ a₂
N- w₂ = 0
N = W₂ = mg
μ₂ mg = m a₂
a = μ₂ g
a = 0.25 9.8
a = 2.45 m / s²
Per calcolare la pressione a una certa profondità, devi considerare la legge di Stevino:
<span>p = ρ · g · h
Tenendo conto che:
g = 9,81 m/s²
ρ = 1000 kg/m³
Troviamo:
p(h</span>₁) = ρ · g · h₁ = 1000 · 9,81 · 4,50 = 44145 Pa
p(h₂) = ρ · g · h₂ = 1000 · 9,81 · 5,50 = 53955 Pa
Answer:
Explanation:
(1.7 m/cycle)(46 cycle/s) = 78.2 m/s
Answer:
Explanation:
The electric flux through the rectangle is given by
where
E is the electric field strength
A is the area of the rectange
is the angle between the direction of the electric field and of the vector normal to the plane of the rectangle
In this problem we have
E = 125 000 N/C
The area of the rectangle is
and the angle is
so, the electric flux is
