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strojnjashka [21]
3 years ago
11

A mango is dropped and fall freely from rest. What are its position and velocity after 1.0secs,2.0secs,and 3.0secs

Physics
1 answer:
irakobra [83]3 years ago
4 0

Answer:

sxsdfsd

Explanation:

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Evaluate the gravitational potential energy between two 5.00-kg spherical steel balls separated by a center-tocenter distance of
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Answer:

U = 8.30×10-⁹J

Explanation:

m1 = m2 = 5.00kg masses of the spheres

d = 15.0cm = 15×10-²m

r = 5.10cm = 5.10×10-²m

R = d + r = 15×10-² + 5.10×10-²

R = 20.10 ×10-²m = 0.201m

G = 6.67×10-¹¹Nm²/kg²

U = Gm1×m2/R = potential energybetween the spheres

U = 6.67×10-¹¹×5.00×5.00/0.201

U = 8.30×10-⁹J

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Una pelota de béisbol de 142g de masa, luego de ser arrojada por el pitcher lleva una velocidad de 90mph. Luego de ser bateada s
V125BC [204]

Answer:

a)    I = 13.38 kg m / s, b)    F = 1,373 10³ N

Explanation:

The impulse is given by the relation

          I = ∫ F dt = Δp

          I = p_f -p₀

          I = m (v_f - v₀)

take the ball's exit direction as positive, whereby the ball velocities

v₀ = -90mph, the final velocity v_f = + 54 m / s

Let's reduce the units to

         I = 0.142 [54- (-40.23) ]

      the SI system

        v₀ = - 90 mph (1609.34 m / 1 mile) (1h / 3600 s = -40.23 m / s

        m = 142 g (1kg / 1000) = 0.142 kg

we calculate  

          I = 0.142 [54- (-40) ]

          I = 13.38 kg m / s

b) let's use the definition of momentum

         I = ∫ F .dt

         I = F ∫ dt

         F = I / t

         F = 13.38 / 0.008

         F = 1,373 10³ N

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