Respuesta :
Answer:
the correct answer is:
D. The gravitational potential energy of the object decreases if [tex]\( d_0 < d \)[/tex]
Explanation:
When the object is submerged in water, the buoyant force acting on the object is equal to the weight of the water displaced by the object. The buoyant force is given by Archimedes' principle:
[tex]\[ F_{\text{buoyant}} = \rho_{\text{water}} \cdot g \cdot V_{\text{displaced}} \][/tex]
where:
[tex]- \( \rho_{\text{water}} \)[/tex] is the density of water,
[tex]- \( g \)[/tex] is the acceleration due to gravity,
[tex]- \( V_{\text{displaced}} \)[/tex] is the volume of water displaced by the submerged object.
The weight of the object is given by:
[tex]\[ F_{\text{weight}} = \rho_{\text{object}} \cdot g \cdot V_{\text{object}} \][/tex]
where:
[tex]- \( \rho_{\text{object}} \)[/tex] is the density of the object,
[tex]- \( V_{\text{object}} \)[/tex] is the volume of the object.
If the object is released and moves a vertical distance [tex]\( h \)[/tex] the water, the work done by the buoyant force is given by:
[tex]\[ W_{\text{buoyant}} = F_{\text{buoyant}} \cdot h \][/tex]
Now, consider the change in gravitational potential energy [tex](\( \Delta U \))[/tex]
[tex]\[ \Delta U = U_{\text{initial}} - U_{\text{final}} \][/tex]
[tex]The gravitational potential energy initially is \( U_{\text{initial}} = m \cdot g \cdot h_{\text{initial}} \), where \( m \) is the mass of the object and \( h_{\text{initial}} \) is the initial height.[/tex]
[tex]The gravitational potential energy finally is \( U_{\text{final}} = m \cdot g \cdot h_{\text{final}} \), where \( h_{\text{final}} \) is the final height.[/tex]
Since [tex]\( W_{\text{buoyant}} = \Delta U \)[/tex]e is equal to the change in gravitational potential energy.
Now, let's analyze the options:
A. The gravitational potential energy of the water in the vessel increases if [tex]\( d_0 < d \)[/tex] It's not about the water's gravitational potential energy.
B. The gravitational potential energy of the water in the vessel decreases if [tex]\( d_0 < d \)[/tex] It's not about the water's gravitational potential energy.
C. The gravitational potential energy of the object increases if [tex]\( d_0 > d \)[/tex]
- Incorrect. The object is moving upward; its gravitational potential energy decreases.
D. The gravitational potential energy of the object decreases if [tex]\( d_0 < d \)[/tex] - Correct. As the object moves upward, its gravitational potential energy decreases.
So, the correct answer is:
D. The gravitational potential energy of the object decreases if [tex]\( d_0 < d \).[/tex]
Final answer:
The correct answer is D) the gravitational potential energy of the object decreases if d₀ < d.
Explanation:
The gravitational potential energy of the object decreases if d₀ < d. When an object of density d₀ is released deep inside water of density d and moves a vertical distance h within the water, its gravitational potential energy decreases. This is because as the object moves upward, its height above the point of reference decreases, resulting in a decrease in gravitational potential energy. Therefore, the correct answer would be D)The gravitational potential energy of the object decreases if d₀ < d.
