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:
(a) 110 m/s
(b) 42 m/s
Explanation:
mass, m = 400 kg, F = 3200 N, V1 = 2 m/s,
acceleration, a = Force / mass = 3200 / 400 = 8 m/s^2
(a) Use first equation of motion
v = V1 + a t
v = 2 + 8 x 1 = 10 m/s
(b) Again using first equation of motion
v = V1 + a t
v = 2 + 8 x 5 = 42 m/s
Thus, the velocity of plane after 1 second is 10 m/s and after 5 second the velocity is 42 m/s.
The erroneous theory of Galileo was that tides were a result of the earth's what he called the inconsistent revolution around the sun. He was right about the most important thing of the earth revolving around the sun but didn't realize that the moon (and sun) exert a gravitational pull on the earth's oceans and cause the diurnal tides to occur.
True, charge can not be created or destroyed, only transferred
<span>Since there is no friction, conservation of energy gives change in energy is zero
Change in energy = 0
Change in KE + Change in PE = 0
1/2 x m x (vf^2 - vi^2) + m x g x (hf-hi) = 0
1/2 x (vf^2 - vi^2) + g x (hf-hi) = 0
(vf^2 - vi^2) = 2 x g x (hi - hf)
Since it starts from rest vi = 0
Vf = squareroot of (2 x g x (hi - hf))
For h1, no hf
Vf = squareroot of (2 x g x (hi - hf))
Vf = squareroot of (2 x 9.81 x 30)
Vf = squareroot of 588.6
Vf = 24.26
For h2
Vf = squareroot of (2 x 9.81 x (30 – 12))
Vf = squareroot of (9.81 x 36)
Vf = squareroot of 353.16
Vf = 18.79
For h3
Vf = squareroot of (2 x 9.81 x (30 – 20))
Vf = squareroot of (20 x 9.81)
Vf = 18.79</span>
