Answer:
Two major causes are outline bellow
1. The presence of air in the system
2. Clogged condenser
Explanation:
1. The presence of air in the system
One of the causes that have been established in relation to high compressor discharge pressure is the presence of air in the system. When this takes place, your best solution is to recharge the system.
2. Clogged condenser
Another is a clogged condenser in which case you will need to clean the condenser so that it will function properly. When you happen to spot that the discharge valve is closed and it is causing high discharge pressure on the compressor, you can solve that easily by opening the valve
Answer:A- mass charge.
This can also be called current.
Explanation:
This is Kirchhoff’s 2nd law.
Kirchhoff’s junction law states that the sum of current(mass charge) flowing in and out of the junction must be equal to zero. This law emphasizes conservation of charge and energy. Charge is also a form of energy and it can neither be created nor destroyed.
Answer:
Broadcasting is the method, not sure about the stage it is done in
Explanation:
Answer:
a) K = 0.63 J, b) h = 0.153 m
Explanation:
a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is
w² =
where d is the distance from the pivot point to the center of mass and I is the moment of inertia.
The rod is a homogeneous body so its center of mass is at the geometric center of the rod.
d = L / 2
the moment of inertia of the rod is the moment of a rod supported at one end
I = ⅓ m L²
we substitute
w =
w =
w =
w = 4.427 rad / s
an oscillatory system is described by the expression
θ = θ₀ cos (wt + Φ)
the angular velocity is
w = dθ /dt
w = - θ₀ w sin (wt + Ф)
In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1
In the exercise it is indicated that at the lowest point the angular velocity is
w = 4.0 rad / s
the kinetic energy is
K = ½ I w²
K = ½ (⅓ m L²) w²
K = 1/6 m L² w²
K = 1/6 0.42 0.75² 4.0²
K = 0.63 J
b) for this part let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ I w²
final point. Highest point
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ I w² = m g h
½ (⅓ m L²) w² = m g h
h = 1/6 L² w² / g
h = 1/6 0.75² 4.0² / 9.8
h = 0.153 m
Answer:
The answer your looking for is option 2 - Inertia