Answer:
(A) 10052.2 m/s²
(B) 0.00678 seconds
Explanation:
From the question,
(A) Applying
V² = U²+2as..................... Equation 1
Where V = Final velocity, U = Initial velocity, a = acceleration due to gravity, s = distance.
make a the subject of the equation
a = (V²-U²)/2s........................ Equation 2
Given: U = 0 m/s( from rest), V = 68 m/s, s = 0.230 m
Substitute these values into equation 2
a = (68²-0²)/(2×0.230)
a = 10052.2 m/s²
(B) Using,
a = (V-U)/t......................... Equation 3
Where t= time.
make t the subject of the equation
t = (V-U)/a......................... Equation 4
Given: V = 68 m/s, U = 0 m/s, a = 10052.2
Substitute into equation 4
t = (68-0)/10052.2
t = 0.00678 seconds
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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Answer:
Explanation:
A.
Given:
Vo = 21 m/s
Vf = 0 m/s
Using equation of Motion,
Vf^2 = Vo^2 - 2aS
S = (21^2)/2 × 9.8
= 22.5 m.
B.
Given:
S = 22.5 + 21 mm
= 22.521 m
Vo = 0 m/s
Using the equation of motion,
S = Vo × t + 1/2 × a × t^2
22.521 = 0 + 1/2 × 9.8 × t^2
t^2 = (2 × 22.521)/9.8
= 4.6
t = 2.14 s
23. 500x10=5,000 joules
24. 500x5=2,500 joules
25. 2,500 joules at point B
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