Answer:
1. increases
2. increases
3. increases
Explanation:
Part 1:
First of all, since the box remains at rest, the horizontal net force acting on the box must equal zero:
F1 - fs = 0.
And this friction force fs is:
fs = Nμs,
where μs is the static coefficient of friction, and N is the normal force.
Originally, the normal force N is equal to mg, where m is the mass of the box, and g is the constant of gravity. Now, there is an additional force F2 acting downward on the box, which means it increases the normal force, since the normal force by Newton's third law, is the force due to the surface acting on the box upward:
N = mg + F2.
So, F2 is increasing, that means fs is increasing too.
Part 2:
As explained in the part 1, N = mg + F2. F2 is increasing, so the normal force is thus increasing.
Part 3:
In part 1 and part 2, we know that fs = Nμs, and since the normal force N is increasing, the maximum possible static friction force fs, max is also increasing.
<span> </span>Most of the stars in the Milky Way will end their lives as white dwarfs
Answer:
2.3 Nm clockwise
Explanation:
Take counterclockwise to be positive and clockwise to be negative.
∑τ = (3 N) (2.5 m) − (7 N) (1.4 m)
∑τ = 7.5 Nm − 9.8 Nm
∑τ = -2.3 Nm
The net torque is 2.3 Nm clockwise.
<h2>
Spring constant is 14.72 N/m</h2>
Explanation:
We have for a spring
Force = Spring constant x Elongation
F = kx
Here force is weight of mass
F = W = mg = 0.54 x 9.81 = 5.3 N
Elongation, x = 36 cm = 0.36 m
Substituting
F = kx
5.3 = k x 0.36
k = 14.72 N/m
Spring constant is 14.72 N/m
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
