Answer:
10.4 m/s
Explanation:
The problem can be solved by using the following SUVAT equation:
![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
For the diver in the problem, we have:
is the initial velocity (positive because it is upward)
is the acceleration of gravity (negative because it is downward)
By substituting t = 1.7 s, we find the velocity when the diver reaches the water:
![v=+6.3 + (-9.8)(1.7)=-10.4 m/s](https://tex.z-dn.net/?f=v%3D%2B6.3%20%2B%20%28-9.8%29%281.7%29%3D-10.4%20m%2Fs)
And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.
The electrons are already there. They are freely moving through the conductor.
The electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.
To determine σ:
σ = Q/A
Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:
σ = Q/d²
Make this substitution in the equation for E:
E = Q/(2ε₀d²)
We see that E is inversely proportional to the square of d:
E ∝ 1/d²
The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude:
![E_{new} = E/4](https://tex.z-dn.net/?f=E_%7Bnew%7D%20%3D%20E%2F4)
<span>The answer is 6 kg the mass of the second
object. By using Inversely proportional formula it means that (14 kg) (3 m/s</span>²<span>)
= M (7 m/s</span>²<span>). Where M is the mass of the second object. For the
Newton’s second law of motion formula which is: Force = mass x acceleration, we
have:</span>
<span>F = (14 kg) (3 m/s</span>²<span>) = 42 N</span>
Therefore:
<span>42 N = M (7 m/s</span>²)
<span>M = (42 N) / (7 m/s</span>²<span>)</span>
M = 6 kg mass of the second object