Answer:
11.8 m/s
Explanation:
At the top of the hill, there are two forces on the car: weight force pulling down (towards the center of the circle), and normal force pushing up (away from the center of the circle).
Sum of forces in the centripetal direction:
∑F = ma
mg − N = m v²/r
At the maximum speed, the normal force is 0.
mg = m v²/r
g = v²/r
v = √(gr)
v = √(9.8 m/s² × 14.2 m)
v = 11.8 m/s
Answer:
a) P =392.4[Pa]; b) F = 706.32[N]
Explanation:
With the input data of the problem we can calculate the area of the tank base
L = length = 10[m]
W = width = 18[cm] = 0.18[m]
A = W * L = 0.18*10
A = 1.8[m^2]
a)
Pressure can be calculated by knowing the density of the water and the height of the water column within the tank which is equal to h:
P = density * g *h
where:
density = 1000[kg/m^3]
g = gravity = 9.81[m/s^2]
h = heigth = 4[cm] = 0.04[m]
P = 1000*9.81*0.04
P = 392.4[Pa]
The force can be easily calculated knowing the relationship between pressure and force:
P = F/A
F = P*A
F = 392.4*1.8
F = 706.32[N]
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
I think it’s b... not sure tho sorry
B. Mentally represent their environment
