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.
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Answer:
dR/dt = 10.2 ft / s
Explanation:
Let's work this problem by finding the distance between the balloon and the motorcycle and then drift for the speed change of the distance
Balloon
y = y₀ +
t
Motorcycle
x = v₀ₓ t
Distance, let's use Pythagoras' theorem
R² = x² + y²
R² = (v₀ₓ t)² + (y₀ +
t)²
v₀ₓ = 88 ft / s
= 8 ft / s
y₀ = 150 ft
R² = (8 t)² + (150 + 8 t )²
R² = 64 t² + (150 + 8t )²
This is the expression for the distance between the two bodies, the rate of change is the derivative with respect to time (d / dt)
2RdR / dt = 64 2 t + 2 (150 + 8t) 8
dR / dt = [64 t + (1200 + 64t )] / R
dR/dt = (1200 +128 t)/R
Let's calculate for the time of 10 s
dR / dt = (1200 + 128 10) / R = 2480 /R
R = √ [64 10² + (150 + 8 10)²
R = √ [6400 + 52900]
R = 243.5 ft
dR / dt = (2480) / 243.5
dR / dt = 10.2 ft / s
Answer:
a) 3 * 10^9 ms
b) 1.6 * 10^-9 gm
c) 6.4 * 10^6 m
d) 5.48 * 10^5 s
Explanation:
Scientific notation is always written in [number from 0-10] (multiplied by) [10] raised to the power 'n'