Answer:
v_f = 24.3 m / s
Explanation:
A) In this exercise there is no friction so energy is conserved.
Starting point. On the roof of the building
Em₀ = K + U = ½ m v₀² + m g y₀
Final point. On the floor
Em_f = K = ½ m v_f²
Emo = Em_g
½ m v₀² + m g y₀ = ½ m v_f²
v_f² = v₀² + 2 g y₀
let's calculate
v_f = √(10² + 2 9.8 25)
v_f = 24.3 m / s
Answer:
To derive the fourth equation of motion, first we have to consider the equation for acceleration and then to rearrange it. or v2 = u2 + 2as and this equation of motion can be used to find the final velocity or the distance travelled if the other values are given.
Explanation:
v= u + at
s =( u + v ) t /2
s = ut + at2/2
v2 = u2 + 2as
Considering the unknown resistence as R and using the Ohm's First Law, we have:
The equivalent resistence is given by the resistor series with the lamp resistence.

If you notice any mistake in my english, please let me know, because i am not native.
Answer:
Since the net force is to the right (in the direction of the applied force), then the applied force must be greater than the friction force. The friction force can be determined using an understanding of net force as the vector sum of all the forces.
Explanation:
The total flux through the cylinder is zero.
In fact, the electric flux through a surface (for a uniform electric field) is given by:

where
E is the intensity of the electric field
A is the surface
is the angle between the direction of E and the perpendicular to the surface, whose direction is always outwards of the surface.
We can ignore the lateral surface of the cylinder, since the electric field is parallel to it, therefore the flux through the lateral surface of the cylinder is zero (because
and
).
On the other two surfaces, the flux is equal and with opposite sign. In fact, on the first surface the flux will be

where r is the radius, and where we have taken
since the perpendicular to the surface is parallel to the direction of the electric field, so
. On the second surface, however, the perpendicular to the surface is opposite to the electric field, so
and
, therefore the flux is

And the net flux through the cylinder is
