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Gelneren [198K]
3 years ago
15

An object is suspended from the ceiling with two wires that make an angle of 40° with the ceiling. The weight of the body is 150

N. What is the tension in each wire?

Physics
2 answers:
dimaraw [331]3 years ago
5 0
With questions involving Forces, it is best to draw a force diagram to model the problem, like the one shown below

We have two tension acting on the particle; T₁ and T₂. The tensions are vectors directed toward the ceiling.
There's also the force as weight acting vertically downwards from the particle. This force is 150N.

As the tension force acting in an angle, we will have to consider the sine and sine components of the tensions. 
We will also solve the force in the direction of x-axis and y-axis

Solving in x-axis direction:
T₂ cos 40° - T₁ cos 40° = 0
T₂ cos 40° = T₁ cos 40°
T₂ = T₁ ...(Eq 1)

Solving in y-axis direction:
T₁ sin 40° + T₂ sin 40° - 150N = 0
T₁ sin 40° + T₂ sin 40° = 150 ⇒ we know from (Eq 1) that T₁ = T₂
2T₁ sin 40° = 150
T₁ sin 40° = 75
T₁ = 75 ÷ sin 40°
T₁ = 116.7 N

Since T₁ = T₂, so T₂ = 116.7N

denpristay [2]3 years ago
5 0

Answer: 117

Explanation:

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Answer:

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Known data

For v_{f} = 25\frac{ft}{s} ,y=\frac{1}{4} h; where h is the maximum height

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Problem development

We replace  v_{f} = 25\frac{ft}{s}  , y=\frac{1}{4} h (ft) in the formula (1),

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0=(v_{i} )^{2} - 2*g*h

2*g*h=(v_{i} )^{2}

h=\frac{(v_{i})^{2}  }{2g}   Equation(2)

We replace (h) of Equation(2) in the  Equation (1) :

25^{2} =(v_{i} )^{2} -2g\frac{\frac{(v_{i})^{2}  }{2g} }{4}

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E = (6.6×10^-34 J s) * 9.85×10^14 / s

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<h3>What is Kepler's third law?</h3>

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