The tension in the rope B is determined as 10.9 N.
<h3>Vertical angle of cable B</h3>
tanθ = (6 - 4)/(5 - 0)
tan θ = (2)/(5)
tan θ = 0.4
θ = arc tan(0.4) = 21.8 ⁰
<h3>Angle between B and C</h3>
θ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰
Apply cosine rule to determine the tension in rope B;
A² = B² + C² - 2BC(cos A)
B = C
A² = B² + B² - (2B²)(cos A)
A² = 2B² - 2B²(cos 43.6)
A² = 0.55B²
B² = A²/0.55
B² = 65.3/0.55
B² = 118.73
B = √(118.73)
B = 10.9 N
Thus, the tension in the rope B is determined as 10.9 N.
Learn more about tension here: brainly.com/question/24994188
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Explanation:
There are five equations of motion:
v = at + v₀
Δx = v₀ t + ½ at²
Δx = ½ (v + v₀)t
v² = v₀² + 2aΔx
Δx = vt − ½ at²
Δx is the displacement
v₀ is the initial velocity
v is the final velocity
a is the acceleration
t is time
Answer:
Explanation: please see attached file I attached the answer to your question.
Answer: If a positively charged ion is more concentrated outside the cell, the forces required to balance the chemical gradient would be directed OUTWARD. Thus, the equilibrium potential for this ion would be POSITIVELY charged. The correct answer is OUTWARD: POSITIVELY.
Explanation: Usually across a cell membrane there is a force that acts on it which is as a result of unequal distribution of charges. This force is known as electrochemical driving force. It is determined by the difference between the membrane potential ( that is, the electrical potential difference across the cell membrane) and the ion equilibrium potential. The membrane potential of a cell helps in signal transmission between different parts of the cell and results when there is unequal distribution across the cell.
Therefore If a positively charged ion is more concentrated outside the cell, the forces required to balance the chemical gradient would be directed outward.Thus, the equilibrium potential for this ion would be positively charged.
Pretty much all other colors except yellow are absorbed.
