Respuesta :
Answer:
T = 34.54 N
Explanation:
First we find the buoyant force acting on the sphere, due to displaced water. For that purpose, we need to find the volume of water displaced by the sphere.
Volume of Water Displaced = V = (Mass of Sphere)/(Density of metal)
V = 4.7 kg/(4000 kg/m³)
V = 0.001175 m³
Now, the buoyant force is given as:
F = (Density of Water)(V)(g)
F = (1000 kg/m³)(0.001175 m³)(9.8 m/s²)
F = 11.52 N
Now, we find the weight of the sphere:
W = mg = (4.7 kg)(9.8 m/s²)
W = 46.06 N
Since, both the tension force (T) and buoyant force act in upward direction, while the weight of sphere act in downward direction. Therefore,
W = T + F
T = W - F
T = 46.06 N - 11.52 N
T = 34.54 N
The tension in the chord when sphere emmerged out of the water is 34.54 N.
Volume of Water Displaced,
[tex]\bold { V = \dfrac m\rho}[/tex]
Where,
m -Mass of Sphere = 4.7 kg
[tex]\bold {\bold { \rho}}[/tex] - Density of metal = 4000 kg/m³
Thus,
[tex]\bold {\bold { V = \dfrac {4.7\ kg}{4000\ kg/m^3}}}\\\\\bold {\bold { V = 0.001175 m^3}}[/tex]
The buoyant force on sphere,
F = (Density of Water) (V) (g)
F = (1000 kg/m³) (0.001175 m³) (9.8 m/s²)
F = 11.52 N
So, weight of the sphere,
W = mg
w = (4.7 kg) (9.8 m/s²)
w = 46.06 N
Since, both the tension force and buoyant force act in upward direction, while the weight of sphere act in opposite direction.
T = W - F
T = 46.06 N - 11.52 N
T = 34.54 N
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