Can somebody help me please

Answer:

yes they can be recycled

Explanation:

Answer:

x₂ / x₁ = √2

Explanation:

To solve this exercise we can use the projectile launch ratios, let's find the time it takes for the second book to reach the ground

             y = y₀ + t - ½ g t²

as the book is thrown horizontally v_{oy} = 0, when it reaches the ground its height is zero y= 0

            0 = y₀ - ½ g t²

            t =

            t = \sqrt{    \frac{2 \ 2h}{ g} }

with this time we calculate the horizontal distance traveled

            x = v₀ t

            x₂ = v₀

now let's calculate the time it takes him to get to the floor when he leaves from the first floor

           t =\sqrt{    \frac{2y_o}{ g} }

the horizontal distance traveled is

           x₁ = v₀

therefore the difference in distance between the two runs is

           Δx = x₂-x₁

           Δx = v₀ \sqrt{ \frac{4h}{g} } - v₀ \sqrt{ \frac{2h}{g} }

            Δx = v₀ \sqrt{ \frac{2h}{g} }    √2

            Δx =√2    x₁

the relationship between the two distances is

             x₂ / x₁ = √2

Answer:

70 N

21°

1.1 m/s²

Explanation:

Draw a free body diagram of the block.  There are three forces:

Weight pulling straight down

Normal force pushing perpendicular to the incline

Friction force pushing parallel to the incline

Part 1

Sum the forces in the perpendicular direction:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

The block is at rest, so F = N μs:

F = N μs

F = mg μs cos θ

F = (20 kg) (9.8 m/s²) (0.38) (cos 19°)

F = 70 N

Part 2

Sum the forces in the parallel direction (down the incline is positive):

∑F = ma

mg sin θ − F = 0

mg sin θ = N μs

mg sin θ = mg μs cos θ

tan θ = μs

θ = atan μs

θ = atan 0.38

θ = 21°

Part 3

Sum the forces in the parallel direction (this time, acceleration is not 0).

∑F = ma

mg sin θ − F = ma

mg sin θ − N μk = ma

mg sin θ − mg μk cos θ = ma

a = g (sin θ − μk cos θ)

a = (9.8 m/s²) (sin 24° − 0.32 cos 24°)

a = 1.1 m/s²