Answer:
B) The entropy is greater in the second state, with the gas on both sides of the box.
Explanation:
As we know that ,this is a irreversible process .The process leaves some effect on the surrounding or on the system itself ,is known as irreversible process.But on the other hand those process does not leave any effect on the system and surrounding is known as reversible process.
The entropy in the irreversible process always increases ,that is why the the entropy will be more when gas occupy the both boxes.
Therefore the answer is --B
Answer:
Explanation:
the center of mass formula
Ycm= [(m₁y₁) + (m₂y₂) + (m₃y₃)] / (m₁+m₂+m₃)
Rope forms the x axis and position of centre of different massses are above or below it so they represent their location on y - axis.
y₁ = 1.6 , y₂ = .7 and y₃ = - 2.1
Ycm ( given ) = - .5
Putting the values of masses and positions
- .5 = 80 x 1.6 + 20 x .7 + m₃ x - 2.1 / ( 80 + 20 + m₃ )
- .5 = 128 + 14 + m₃ x - 2.1 / ( 100+ m₃ )
- 50 - .5 m₃ = 142 - 2.1 m₃
1.6 m₃ = 192
m₃ = 120 kg .
B )
Total downward force is weight of total mass = 80 + 20 + 120
= 220 kg
weight = 220 x 9.8 = 2156 N .
component of weight perpendicular to rope
= 2156 cos 15 = 2082.53 N
This force will be equally distributed over each tree , so force on each tree = 2082.53 / 2 = 1041.26 N .
Answer:
static
Explanation:
static friction pushes in the direction you are walking.
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
An electron that is far away from the nucleus have higher energy than an electron near the nucleus. Nucleus are positively charged and those electrons near it get attracted; those electrons gain kinetic energy hence reducing their internal energy. The electrons far from nucleus have low kinetic energy hence more internal energy.
