energy is the correct answer to fill the blank bb :)
Answer:
The initial velocity was U=22.14m/s
Explanation:
Step one :
Applying the third equation of motion
v² = u²+ 2as
Where v= Final velocity
U =initial velocity
a= acceleration due to gravity
S= distance or displacement
Step two :
V= 0
a= 9.81m/s²
S=25m
U=?
Step three :
Substituting into the equation we have
0²=U²+2*9.81*25
0=U²+490.5
U²=-490.5
U=√490.5
U=22.14m/s
Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:

- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:

- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
When the diver reaches maximum height, the upward velocity will be zero.
We shall use the formula
v^2 = u^2 - 2gh
where
v = 0 (velocity at maximum height)
u = 1.2 m/s, intial upward velocity
g = -9.8 m/s^2, gravitational acceleration (downward)
h = maximum height attained above the diving board.
Therefore
0 = 1.2^2 - 2*9.8*h
h = 1.2^2/(2*9.8) = 0.0735 m
Answer: 0.074 m (nearest thousandth)