It can be described as a constant variation
Answer: 580 N
Refer to attached figure.
The angle of inclination is 22 degrees
weight (gravitational force) acts downwards.
Normal force is a contact force which acts perpendicular to the point of contact.
The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.
Gravitational force on an object = mg
The normal force 
the equation of the tangent line must be passed on a point A (a,b) and
perpendicular to the radius of the circle. <span>
I will take an example for a clear explanation:
let x² + y² = 4 is the equation of the circle,
its center is C(0,0). And we assume that the tangent line passes to the point
A(2.3).
</span>since the tangent passes to the A(2,3), the line must be perpendicular to the radius of the circle.
<span>Let's find the equation of the line parallel to the radius.</span>
<span>The line passes to the A(2,3) and C (0,0). y= ax+b is the standard form of the equation. AC(-2, -3) is a vector parallel to CM(x, y).</span>
det(AC, CM)= -2y +3x =0, is the equation of the line // to the radius.
let's find the equation of the line perpendicular to this previous line.
let M a point which lies on the line. so MA.AC=0 (scalar product),
it is (2-x, 3-y) . (-2, -3)= -4+4x + -9+3y=4x +3y -13=0 is the equation of tangent
If 56.5kJ are needed to raise the temp by 90°C and if the heater is 60% efficient that means that:
60% X y = 56.5kJ
where y is the electrical energy in kJ that the heater will use.
y = 94.2kJ
Answer:
0.368 cm
Explanation:
x = distance by which the mercury rise
d = depth of the water = 10 cm = 0.10 m
ρ = density of water = 1000 kgm⁻³
ρ' = density of mercury = 13600 kgm⁻³
P₀ = atmospheric pressure
Using equilibrium of pressure on both side
P₀ + ρ g d = P₀ + ρ' g (2x)
(1000) (0.10) = (13600) (2x)
x = 0.00368 m
x = 0.368 cm
