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:
243 N
Explanation:
The formula for electromagnetic force is F= Kq1q2/r^2
where r is the distance between the charges, if the distance between the charges is reduced by 1/3 then F will increase by 9 [(1/3r)^2 becomes 1/9r which is 9F] so 27*9 is 243N
Answer: the two states that are fluid are;-
<u>#{1} liquid</u>
<u>#{2} gas </u>
Explanation:
as we know that there are mainly three states of substance
but among them only two of them can fluid and takes the shape of the container that are liquid and gas
1. Pick a point on the top of the object and draw two incident rays traveling towards the mirror. Using a straight edge, accurately draw one ray so that it passes exactly through the focal point on the way to the mirror. Draw the second ray such that it travels exactly parallel to the principal axis.
Answer:
1.5 m
Explanation:
Length. L = 12 m
Width, W = 16 m
Area, A = 12 x 16 = 192 m^2
Let the width of pavement be d.
The new length, L' = 12 + 2d
the new width, W' = 16 + 2d
New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2
Difference in area = A' - A
285 = 192 + 56 d + 4d^2 - 192
93 = 56 d + 4d^2
4d^2 + 56 d - 93 = 0
\
d = 1.5 m
Thus, the width of the pavement is 1.5 m.
