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:
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Explanation:
Answer:
Explanation:
Given:
- temperature of skin,
- initial temperature of steam vapour,
- latent heat of steam,
- mass of steam,
- specific heat of water,
- final temperature,
<em>Assuming that no heat is lost in the surrounding.</em>
<u>We know:</u>
<u>Now the total heat given by the steam to form water at the given conditions:</u>
..............................(1)
where:
latent heat given out by vapour to form water of 100°C
heat given by water of 100°C to come at 34°C.
putting respective values in eq. (1)
is the heat transferred to the skin.
Answer:
it look the same just to tell you
Answer:
a)1815Joules b) 185Joules
Explanation:
Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;
F = ke where;
F is the applied force
k is the elastic constant
e is the extension of the material
From the formula, k = F/e
F1/e1 = F2/e2
If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;
k = 60/0.5
k = 120N/m
a) To get the work done in stretching the spring 5.5m from its position,
Work done by the spring = 1/2ke²
Given k = 120N/m, e = 5.5m
Work done = 1/2×120×5.5²
Work done = 60× 5.5²
Work done = 1815Joules
b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;
Work done = 1/2ke²
Work done =1/2× 120×1.5²
Works done = 60×1.5²
Work done = 135Joules
