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Tresset [83]
3 years ago
12

Which object would have a LARGER gravitational force acting upon it? (assume the objects are at the same height above the Earth.

)
Physics
2 answers:
Solnce55 [7]3 years ago
8 0

Newtons Second law

Force = mass x acceleration

Weight = mass x g

  • weight : the force that earth applies on an object
  • mass : the mass of the object
  • g : gravitational acceleration. Almost constant, around 9.81m/s^2 at the sea level. constant for given heights.

Given the above, the <u>more massive the object is, the larger the force acting upon it. </u>

THAT DOES NOT MEAN THAT IT'l FALL FASTER since they both have the same acceleration (g). Just larger force to the larger mass.

djverab [1.8K]3 years ago
7 0
A) 5 kg block of wood
B) 100 kg person
C) 412 kg motorcycle
D) 2540 kg elephant
Answer: D
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In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue
Andru [333]

Answer:

a)  p₀ = 1.2 kg m / s,  b) p_f = 1.2 kg m / s,  c)   θ = 12.36, d)  v_{2f} = 1.278 m/s

Explanation:

For this exercise we define a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved

 

a) the initial impulse is

         p₀ = m v₁₀ + 0

         p₀ = 0.6 2

         p₀ = 1.2 kg m / s

b) as the system is isolated, the moment is conserved so

        p_f = 1.2 kg m / s

we define a reference system where the x-axis coincides with the initial movement of the cue ball

 

we write the final moment for each axis

X axis

         p₀ₓ = 1.2 kg m / s

         p_{fx} = m v1f cos 20 + m v2f cos θ

         p₀ = p_f

        1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ

         1.2482 = v_{2f} cos θ

Y axis  

        p_{oy} = 0

        p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ

        0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ

        0.2736 = v_{2f} sin θ

we write our system of equations

         0.2736 = v_{2f} sin θ

         1.2482 = v_{2f} cos θ

divide to solve

         0.219 = tan θ

          θ = tan⁻¹ 0.21919

          θ = 12.36

let's look for speed

            0.2736 = v_{2f} sin θ

             v_{2f} = 0.2736 / sin 12.36

            v_{2f} = 1.278 m / s

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Explanation:

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Answer:

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So, 0.699 L of the fluid will overflow

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