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irga5000 [103]
2 years ago
9

A 100. N block sits on a rough horizontal floor. The coefficient of sliding friction between the block and the floor is 0.250. A

horizontal force of 90.0 N acts on the block for 3.00 seconds. Calculate the velocity of the block after 3.00 seconds if it starts from rest.
Physics
1 answer:
zaharov [31]2 years ago
8 0
     Using the Impulse Theorem, we have:

F_{r}.\Delta t=m.\Delta v \\ (F-F_{at}).(3-0)= \frac{P}{g} .(v-0) \\ (90-100.0,25).3= \frac{100v}{10}  \\ v= \frac{3.65}{10}  \\ \boxed {v=19,5 m/s}

If you notice any mistake in English, please let me know, because I'm not native.
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PSYCHO15rus [73]

Answer:

  a) T= 72.9 Nm

  b)      \Theta = 5 \textdegree

Explanation:

From the question we are told that

Mass 85kg

Speed 15m/s

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Cylinder 50 cm wide and 90 cm long centered 1.2 m above the ground

a)Generally equation for force is given by

   F = 0.5 p A v^2

Mathematically solving for force exacted

   F = 0.5*1.2*0.9*0.5*15^2

   F= 60.75 N

Mathematically solving for torque

Torque,

    T = r * F

    T=1.2*60.75

   T= 72.9 Nm

b)Generally in solving for \theta

   Tan\theta = Torque/(Mass * Gravity)

   Tan\theta = (72.9)/(85 * 9.8)

     \Theta = 5 \textdegree

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When is a body said to be motion (movement )?
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A body is said to be in motion when it's position changes continuously.

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5)List an appropriate SI base unit (with a prefix as needed ) for the following:
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B) kg
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spaceship of mass m travels from the Earth to the Moon along a line that passes through the center of the Earth and the center o
satela [25.4K]

Answer:

the correct result is r = 3.71 10⁸ m

Explanation:

For this exercise we will use the law of universal gravitation

          F = - \frac{m_{1} m_{2} }{r^2}

We call the masses of the Earth M, the masses of the moon m and the masses of the rocket m ', let's set a reference system in the center of the Earth, the distance from the Earth to the moon is d = 3.84 108 m

rocket force -Earth

          F₁ = - \frac{m' M }{r^2}

rocket force - Moon

          F₂ = - \frac{m' m }{(d-r)^2}

in the problem ask for what point the force has the relation

          2 F₁ = F₂

let's substitute

          2 2 \frac{M}{r^2} = \frac{m}{(d-r)^2}

          (d-r) ² = \frac{m}{2M} r²

           d² - 2rd + r² = \frac{m}{2M} r²

           r² (1 -\frac{m}{2M}) - 2rd + d² = 0

Let's solve this quadratic equation to find the distance r, let's call

           a = 1 - \frac{m}{2M}

           a = 1 - \frac{7.36 10^{22} }{2 \  5398 10^{24}} = 1 - 6.15 10⁻³

           a = 0.99385

         

            a r² - 2d r + d² = 0

           r =  \frac  {2d \frac{+}{-}   \sqrt{4d^2 - 4 a d^2}} {2a}

           r = [2d ± 2d \sqrt{1-a}] / 2a

           r = \frac{d}{a}   (1 ± √ (1.65 10⁻³)) =  \frac{d}{a} (1 ± 0.04)

           r₁ = \frac{d}{a} 1.04

           r₂ = \frac{d}{a} 0.96

let's calculate

           r₁ = \frac{3.84 10^8}{0.99385} 1.04

           r₁ = 401.8 10⁸ m

          r₂ = \frac{3.84 10^8}{0.99385} 0.96

          r₂ = 3.71 10⁸ m

therefore the correct result is r = 3.71 10⁸ m

3 0
2 years ago
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