Sb-28 a collision could occur when the distance decreases and bearing between two vessels does what?
1 answer:
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Answer:
on increasing pressure, temperature will also increase.
Explanation:
Considering the ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Thus, at constant volume and number of moles, Pressure of the gas is directly proportional to the temperature of the gas.
P ∝ T
Also,
Also, using Gay-Lussac's law,
Thus, on increasing pressure, temperature will also increase.
Answer:
The weight of the object X is approximately 3.262 N (Acting downwards)
The weight of the object Y is approximately 8.733 N (Acting downwards)
Explanation:
The question can be answered based on the principle of equilibrium of forces
The given parameters are;
The weight of Z = 12 N (Acting downwards)
The weight of the pulleys = Negligible
From the diagram;
The tension in the in the string attached to object Z = The weight of object Z = 12 N
The tension in the in the string attached to object X = The weight of the object X
The tension in the in the string attached to object Y = The weight of the object Y
Given that the forces are in equilibrium, we have;
The sum of vertical forces acting at a point, = 0
Therefore;
Where;
= The weight of object Z = 12 N
= 12 N
= The vertical component of tension, T₂ = T₂ × sin(24°)
∴ = T₂ × sin(156°)
Similarly;
= T₃ × sin(50°)
From , and
= 12 N, we have;
12 N = -(T₂ × sin(156°) + T₃ × sin(50°))...(1)
Given that the forces are in equilibrium, we also have that the sum of vertical forces acting at a point, ∑Fₓ = 0
Therefore at point B, we have;
T₁ₓ + T₂ₓ + T₃ₓ = 0
The tension force, T₁, only has a vertical component, therefore;
∴ T₁ₓ = 0
∴ T₂ₓ + T₃ₓ = 0
T₂ₓ = -T₃ₓ
T₂ₓ = T₂ × cos(156°)
T₃ₓ = T₃ × cos(50°)
From T₂ₓ = -T₃ₓ, we have;
T₂ × cos(156°) = - T₃ × cos(50°)...(2)
Making T₃ the subject of equation (1) and (2) gives;
Making T₃ the subject of equation in equation (1), we get;
12 = -(T₂ × sin(156°) + T₃ × sin(50°))
∴ T₃ = (-12 - T₂ × sin(156°))/(sin(50°))
Making T₃ the subject of equation in equation (2), we get;
T₂ × cos(156°) = - T₃ × cos(50°)
∴ T₃ = T₂ × cos(156°)/(-cos(50°))
Equating both values of T₃ gives;
(-12 - T₂ × sin(156°))/(sin(50°)) = T₂ × cos(156°)/(-cos(50°))
-12/(sin(50°)) = T₂ × cos(156°)/(-cos(50°)) + T₂ × sin(156°)/(sin(50°))
∴ T₂ = -12/(sin(50°))/((cos(156°)/(-cos(50°)) + sin(156°)/(sin(50°))) ≈ -8.02429905283
∴ T₂ ≈ -8.02 N
From T₃ = T₂ × cos(156°)/(-cos(50°)), we have;
T₃ = -8.02× cos(156°)/(-cos(50°)) = -11.3982199717
∴ T₃ ≈ -11.4 N
The weight of the object X = -T₂ × sin(156°)
∴ The weight of the object X ≈ -(-8.02 × sin(156°)) = 3.262 N
The weight of the object X ≈ 3.262 N (Acting downwards)
The weight of the object Y = -(T₃ × sin(50°))
∴ The weight of the object Y = -(-11.4 × sin(50°)) ≈ 8.733 N
The weight of the object Y ≈ 8.733 N (Acting downwards)
