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Natali5045456 [20]
3 years ago
15

What is the kinetic energy of a 0.135 kg baseball thrown at 40 m/s

Physics
1 answer:
Ronch [10]3 years ago
4 0
KE = 1/2 * m * v^2
KE = 1/2 * 0.135 * 40^2
KE = 1/2 * 0.135 * 1600
KE = 108 J
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La ecuación de la posición de una esferita está dada por: r(t)=(2.Cos(πt) i-3.Sen(πt) j) (m) ¿Cuál es la velocidad de la esferit
katovenus [111]

Answer:

v = (-4.44 i^ + 6.66 j^ )  m/s, a_average =( 0 i^ -2π j^) m/s²

Explanation:

The expression left corresponds to an oscillatory movement (MAS), the speed is defined by

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the function of position

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let's perform the derivative

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          v = -2π sin (π 0.25) i^ + 3π cos (π 0.25) j^

           v = (-4.44 i^ + 6.66 j^ )  m/s

To calculate the mean acceleration we use the expression

           a = (v_{f} - v_{o}) / Δt

 

indicates that the time is the first 3 s

       

we look for the initial velocity t = 0 s

           v₀ = 0 i ^ + 3π j ^

         

we look for the fine velocity, t = 3 s

          v_f = - 2π sin (π 3) + 3π cos (π 3) j ^

          v_f = 0 i ^ - 3π j ^

we calculate the average acceleration

            Δt = (3 -0) = 3 s

           a_average = (0-0) / 3 i ^ + (-3π - 3π) / 3

           a_average = (0 i ^ -2π j ^ ) m/s²

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3 years ago
PLEASE HELP!!!! The displacement vectors A and B, when added together, give the resultant vector R, so that R = A + B. Use the d
GuDViN [60]

Answer:

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3 years ago
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Ahat [919]
The answer above me is correct
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3 years ago
Does the KE of a car change more when it accelerates from 22 km/h to 32 km/h or when it accelerates from 32 km/h to 42 km/h
mote1985 [20]

Answer:

The change in kinetic energy (KE) of the car is more in the second case.

Explanation:

Let the mass of the car = m

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change in kinetic energy (K.E) = ¹/₂m(v² - u²)

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initial velocity of the second case, u = 32 km/h = 8.89 m/s

final velocity of the second case, v = 42 km/h = 11.67 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

                                          ΔK.E = ¹/₂m(11.67² - 8.89²)

                                                    = 28.58m J

The change in kinetic energy (KE) of the car is more in the second case.

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