This can be answered using trigonometric analysis. This sloped path that is 150 m long is the hypotenuse of the triangle. The adjacent angle would then be 65 degrees. Given these:
sin 65 = h / 150
Where: h = vertical displacement = 150 (sin 65)
h = 135.95 meters
The magnitudes of the forces that the ropes must exert on the knot connecting are :
- F₁ = 118 N
- F₂ = 89.21 N
- F₃ = 57.28 N
<u>Given data :</u>
Mass ( M ) = 12 kg
∅₂ = 63°
∅₃ = 45°
<h3>Determine the magnitudes of the forces exerted by the ropes on the connecting knot</h3><h3 />
a) Force exerted by the first rope = weight of rope
∴ F₁ = mg
= 12 * 9.81 ≈ 118 kg
<u>b) Force exerted by the second rope </u>
applying equilibrium condition of force in the vertical direction
F₂ sin∅₂ + F₃ sin∅₃ - mg = 0 ---- ( 1 )
where: F₃ = ( F₂ cos∅₂ / cos∅₃ ) --- ( 2 ) applying equilibrium condition of force in the horizontal direction
Back to equation ( 1 )
F₂ = [ ( mg / cos∅₂ ) / tan∅₂ + tan∅₃ ]
= [ ( 118 / cos 63° ) / ( tan 63° + tan 45° ) ]
= 89.21 N
<u />
<u>C ) </u><u>Force </u><u>exerted by the</u><u> third rope </u>
Applying equation ( 2 )
F₃ = ( F₂ cos∅₂ / cos∅₃ )
= ( 89.21 * cos 63 / cos 45 )
= 57.28 N
Hence we can conclude that The magnitudes of the forces that the ropes must exert on the knot connecting are :
F₁ = 118 N, F₂ = 89.21 N, F₃ = 57.28 N
Learn more about static equilibrium : brainly.com/question/2952156
Answer:
hey what is this 2 in between the question
please tell
The two main function of a transistor is to amplify or switch electronic signals. The switch is two options: on / off. The amplification function is a change in voltage.
Answer:
a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀
Explanation:
The speed of a wave along an eta string given by the expression
v = 
where T is the tension of the string and μ is linear density
a) the mass of the cable is double
m = 2m₀
let's find the new linear density
μ = m / l
iinitial density
μ₀ = m₀ / l
final density
μ = 2m₀ / lo
μ = 2 μ₀
we substitute in the equation for the velocity
initial v₀ = 
with the new dough
v = 
v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }
v = 1 /√2 v₀
v = 0.7071 v₀
b) we double the length of the cable
If the cable also increases its mass, the relationship is maintained
μ = μ₀
in this case the speed does not change
c) the cable l = l₀ and m = 3m₀
we look for the density
μ = 3m₀ / l₀
μ = 3 m₀/l₀
μ = 3 μ₀
v = 
v = 1 /√3 v₀
v = 0.577 v₀
d) l = 2l₀
μ = m₀ / 2l₀
μ = μ₀/ 2
v = 
v = √2 v₀
v = 1.41 v₀
e) m = 10m₀ and l = 2l₀
we look for the density
μ = 10 m₀/2l₀
μ = 5 μ₀
we look for speed
v = 
v = 1 /√5 v₀
v = 0.447 v₀
