If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her feet are 70% of the way up the length of the ladder, what is the coefficient of static friction between the ladder and the floor

Answer:

b

Explanation:

Answer:

the direction of rate of change of the momentum is against the motion of the body, that is, downward.

The applied force is also against the direction of motion of the body, downward.

Explanation:

The change in the momentum of a body, if the mass of the body is constant, is given by the following formula:

[tex]\Delta p=\Delta (mv)\\\\\Delta p=m\Delta v[/tex]

p: momentum

m: mass

[tex]\Delta v[/tex]:  change in the velocity

The sign of the change in the velocity determines the direction of rate of change. Then you have:

[tex]\Delta v=v_2-v_1[/tex]

v2: final velocity = 35m/s

v1: initial velocity = 40m/s

[tex]\Delta v =35m/s-40m/s=-5m/s[/tex]

Hence, the direction of rate of change of the momentum is against the motion of the body, that is, downward.

The applied force is also against the direction of motion of the body, downward.

Answer:

The velocity is  [tex]v = 8.85 m/s[/tex]

Explanation:

From the question we are told that

    The mass of the skier is [tex]m_s = 60 \ kg[/tex]

      The initial speed is [tex]u = 4.0 \ m/s[/tex]

       The height is  [tex]h = 10 \ m[/tex]

According to the law of energy conservation

     [tex]PE_t + KE_t = KE_b + PE_b[/tex]

Where [tex]PE_t[/tex] is the potential energy at the top which is mathematically evaluated as

       [tex]PE_t = mg h[/tex]

substituting values

       [tex]PE_t = 60 * 4*9.8[/tex]

      [tex]PE_t = 2352 \ J[/tex]

And  [tex]KE_t[/tex] is the kinetic energy at the top which equal to zero due to the fact that velocity is zero at the top of the hill

And  [tex]KE_b[/tex] is the kinetic energy at the bottom of the hill which is mathematically represented as

         [tex]KE_b = 0.5 * m * v^2[/tex]

  substituting  values

         [tex]KE_b = 0.5 * 60 * v^2[/tex]

=>     [tex]KE_b = 30 v^2[/tex]

Where v is the velocity at the bottom

   And [tex]PE_b[/tex] is the potential  energy at the bottom which equal to zero due to the fact that height  is zero at the bottom of the hill

So  

        [tex]30 v^2 = 2352[/tex]

=>      [tex]v^2 = \frac{2352}{30}[/tex]

=>       [tex]v = \sqrt{ \frac{2352}{30}}[/tex]

        [tex]v = 8.85 m/s[/tex]

Answer:

The Skier's velocity at the bottom of the hill will be 18m/s

Explanation:

This is simply the case of energy conversion between potential and kinetic energy. Her potential energy at the top of the hill gets converted to the kinetic energy she experiences at the bottom.

That is

[tex]mgh = 0.5 mv^{2}[/tex]

solving for velocity, we will have

[tex]v= \sqrt{2gh}[/tex]

hence her velocity will be

[tex]v=\sqrt{2 \times 9.81 \times 10}=14.00m/s[/tex]

This is the velocity she gains from the slope.

Recall that she already has an initial velocity of 4m/s. It is important to note that since velocities are vector quantities, they can easily be added algebraically. Hence, her velocity at the bottom of the hill is 4 + 14 = 18m/s

The Skier's velocity at the bottom of the hill will be 18m/s