A stationary object with a mass of 4.4 kg is hit with an average force of 83.7 N. The interaction time $\Delta t$ is 147 milliseconds. Calculate the final speed of the object in m/s, with two digits of precision.

The final speed of the object of mass 4.4 Kg when hit with an average force of 83.7 N is approximately 2.82 m/s (rounded to two digits of precision).

To calculate the final speed of the object, we can use the impulse-momentum theorem, which states that the change in momentum of an object is equal to the impulse applied to it.

Given:

Mass of the object (m) = 4.4 kg

Average force applied (F) = 83.7 N

Interaction time (Δt) = 147 milliseconds = 147 ms = 0.147 s (converted to seconds)

Impulse (J) = F × Δt

Now, the change in momentum (Δp) is given by the impulse:

Δp = J = F × Δt

The final speed (v_f) of the object can be calculated using the equation for momentum (p = m × v):

Δp = m × v_f - 0 (assuming the object starts from rest)

Now, rearrange the equation to solve for the final speed:

v_f = Δp / m

Substitute the values:

v_f = (F × Δt) / m

v_f = (83.7 N × 0.147 s) / 4.4 kg

v_f ≈ 2.8182 m/s

A stationary object with a mass of 4.4 kg is hit with an average force of 83.7 N. The interaction time \(\Delta t\) is 147 milliseconds. Calculate the final speed of the object in m/s, with two digits of precision.

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