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miss Akunina [59]
2 years ago
14

URGENT HELP !! The coefficients of static and kinetic frictions for plastic on wood are 0.53 and 0.40, respectively. How much ho

rizontal force would you need to apply to a 34.4 kg object to start it moving from rest?
Physics
1 answer:
IRISSAK [1]2 years ago
7 0

Answer:

43.83 N

Explanation:

Given that,

The mass of an object, m = 34.4 kg

The coefficients of static and kinetic frictions for plastic on wood are 0.53 and 0.40, respectively.

The force of static friction,

F_s=\mu_smg\\\\F_s=0.53\times 34.4\times 9.8\\\\F_s=178.67\ N

The force of kinetic friction,

F_k=\mu_kmg\\\\F_k=0.40\times 34.4\times 9.8\\\\F_k=134.84\ N

Net force acting on the object is :

F = 178.67-134.84

= 43.83  N

Hence, this is the required solution.

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The main reason is that very young calves are more noticeable to predators when mixed with older calves from the previous year

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A commuter airplane starts from an airport and takes theroute. The plane first flies to city A, located 175 km away in a directi
bearhunter [10]

Answer:

245.45km in a direction 21.45° west of north from city A

Explanation:

Let's place the origin of a coordinate system at city A.

The final position of the airplane is given by:

rf = ra + rb + rc    where ra, rb and rc are the vectors of the relative displacements the airplane has made. If we separate this equation into its x and y coordinates:

rfX = raX+ rbX + rcX = 175*cos(30)-150*sin(20)-190 = -89.75km

rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km

The module of this position is:

rf = \sqrt{rfX^2+rfY^2} = 245.45km

And the angle measure from the y-axis is:

\alpha =atan(rfX/rfY) = 21.45\°

So the answer is 245.45km in a direction 21.45° west of north from city A

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How does the suns gravitational attraction impact earths motion
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Why metals have thermoconductivity higher than ceramic?
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The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0
katen-ka-za [31]

Answer:

a)  F = 2.66 10⁴ N, b)   h = 1.55 m

Explanation:

For this fluid exercise we use that the pressure at the tap point is

Exterior

          P₂ = P₀ = 1.01 105 Pa

inside

         P₁ = P₀ + ρ g h

the liquid is water with a density of ρ=1000 km / m³

         P₁ = 0.85   1.01 10⁵ + 1000   9.8  5

         P₁ = 85850 + 49000

         P₁ = 1.3485 10⁵ Pa

the net force is

         ΔP = P₁- P₂

         Δp = 1.3485 10⁵ - 1.01 10⁵

         ΔP = 3.385 10⁴ Pa

Let's use the definition of pressure

         P = Fe / A

         F = P A

the area of ​​a circle is

         A = pi r² = [i d ^ 2/4

let's reduce the units to the SI system

         d = 100 cm (1 m / 100 cm) = 1 m

         F = 3.385 104 pi / 4 (1) ²

         F = 2.66 10⁴ N

b) the height for which the pressures are in equilibrium is

        P₁ = P₂

        0.85 P₀ + ρ g h = P₀

        h = \frac{P_o ( 1-0.850)}{\rho \ g}

        h = \frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}

        h = 1.55 m

4 0
2 years ago
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