Not sure what you mean by "breaks in the tension" but I suspect you mean the rope will come apart if the tension in the rope exceeds 1800 N.
In the free body diagram for the 500 N weight, we have a figure Y with the net force equations
• horizontal net force:
∑ F[hor] = T₁ cos(θ) - T₂ cos(θ) = 0
• vertical net force:
∑ F[ver] = T₁ sin(θ) + T₂ sin(θ) - 500 N = 0
From the first equation, it follows that T₁ = T₂, so I'll denote their magnitude by T alone. From the second equation, we have
2 T sin(θ) = 500 N
and if the maximum permissible tension is T = 1800 N, it follows that
sin(θ) = (500 N) / (3600 N) ⇒ θ = arcsin(5/36) ≈ 7.9°
is the smallest angle the rope can make with the horizontal.
Answer:
Explanation:
Angular speed of hoop ω = v / r
= 8.90 / .27
= 32.96 rad / s
Translational kinetic energy = 1/2 mv²
= .5 x 1.8 x 8.9²
= 71.29 J
Rotational kinetic energy = 1/2 Iω²
= 1/2 mR²x ω²
= 1/2 mv²
= 71.29 J
Total kinetic energy
= 2 x 71.29
= 142.58 J
This energy will be used to attain height
If h be the height attained
mgh = 142.58
h = 142.58 / mg
= 142.58 / 1.8 x 9.8
= 8.08 m .
<u>Answer:</u>
<em>The shear modulus of the cube material is
.
</em>
<u>Explanation:</u>
<em>Given that shearing force applied F = 1500 N </em>
<em>Displacement produced x = 0.1 cm=0.001 m </em>
<em>side of the cube =20 cm = 0.2 m
</em>
Since the object is a cube the upper surface is a square and it is on this surface the shearing
force is applied
<em>area of the upper surface
</em>
<em>shear strain = tan θ =
</em>
<em>shearing stress =
</em>
<em>modulus of rigidity η
</em>
<em>
</em>