How much thermal energy is needed to melt 1.25 kg of water at its melting point? Use Q = masslaten heat of fusion.
1 answer:
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Answer:
T₂ = 123.9 N, θ = 66.2º
Explanation:
To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.
The tension T1 = 100 N, we create a reference frame centered on the pole
X axis
T₁ₓ - = 0
T_{2x}= T₁ₓ
Y axis y
T_{1y} + T_{2y} - 200N = 0
T_{2y} = 200 -T_{1y}
let's use trigonometry to find the component of the stresses
sin 60 = T_{1y} / T₁
cos 60 = t₁ₓ / T₁
T_{1y} = T₁ sin 60
T1x = T₁ cos 60
T_{1y}y = 100 sin 60 = 86.6 N
T₁ₓ = 100 cos 60 = 50 N
for voltage 2 it is done in the same way
T_{2y} = T₂ sin θ
T₂ₓ = T₂ cos θ
we substitute
T₂ sin θ= 200 - 86.6 = 113.4
T₂ cos θ = 50 (1)
to solve the system we divide the two equations
tan θ = 113.4 / 50
θ = tan⁻¹ 2,268
θ = 66.2º
we caption in equation 1
T₂ cos 66.2 = 50
T₂ = 50 / cos 66.2
T₂ = 123.9 N
Explanation:
Given that,
Work done to stretch the spring, W = 130 J
Distance, x = 0.1 m
(a) We know that work done in stretching the spring is as follows :
(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m
So,
So, the new work is more than 130 J.