Mechanical energy E = mgh + 1/2mv²
When he starts, let h = 0 ⇒ E₁ = 1/2mv₁²
When he reaches height h ⇒ E₂ = mgh + 1/2mv₂²
Without friction, energy is conserved at all times.
E₁ = E₂
↓
1/2mv₁² = mgh + 1/2mv₂²
↓
1/2v₁² = gh + 1/2v₂²
↓
gh = 1/2(v₁² - v₂²)
↓
h = (v₁² - v₂²) / (2g)
Answer:
The sphere C carries no net charge.
Explanation:
- When brougth close to the charged sphere A, as charges can move freely in a conductor, a charge equal and opposite to the one on the sphere A, appears on the sphere B surface facing to the sphere A.
- As sphere B must remain neutral (due to the principle of conservation of charge) an equal charge, but of opposite sign, goes to the surface also, on the opposite part of the sphere.
- If sphere A is removed, a charge movement happens in the sphere B, in such a way, that no net charge remains on the surface.
- If in such state, if the sphere B (assumed again uncharged completely, without any local charges on the surface), is touched by an initially uncharged sphere C, due to the conservation of charge principle, no net charge can be built on sphere C.
Answer:
5.886 J
Explanation:
Given:
The mass of the book is,
kg
Height of lift is,
m
Acceleration due to gravity is,
m/s²
Now, gain in gravitational potential energy is a function of change in position and is given as:

Here,
is the change in gravitational potential energy.
Plug in 0.5 kg for
, 9.81 for
and 1.2 for
. Solve for 

Therefore, the gain in gravitational energy of the book is 5.886 J.
The answer i got was c. gravity and strong
i hope it helps