vf = 10 m/s. A ball with mass of 4kg and a impulse given of 28N.s with a intial velocity of 3m/s would have a final velocity of 10 m/s.
The key to solve this problem is using the equation I = F.Δt = m.Δv, Δv = vf - vi.
The impulse given to the ball with mass 4Kg is 28 N.s. If the ball were already moving at 3 m/s, to calculate its final velocity:
I = m(vf - vi) -------> I = m.vf - m.vi ------> vf = (I + m.vi)/m ------> vf = I/m + vi
Where I 28 N.s, m = 4 Kg, and vi = 3 m/s
vf = (28N.s/4kg) + 3m/s = 7m/s + 3m/s
vf = 10 m/s.
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Answer:
I would say the answer is the wave of 21.000Hz
Explanation:
Because it has more frequency, and as more frequency you add, the time or longer period also increases.
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The breaking distance consists of two parts. The first part is the first 0.5 seconds were no breaking occurs. Given values: t time, v₀ initial velocity:
x₁ = v₀*t
The second part occurs after t = 0,5s with the given acceleration: a = - 12 m/s²
were the final velocity is zero, v = 0 and the initial velocity v₀= 16m/s:
v = a*t + v₀ = 0 => v₀ = -a*t => t = v₀/-a
x₂ = 0.5*a*t² = 0.5*v°²/a
The total breaking distance is the sum of the two parts:
x = x₁ + x₂ = v₀* t + 0.5 * v₀² / a = 16 * 0.5 + 0.5 * 16² / 12 = 8 + 10,7 = 18,7
You can use this result to calculate the remaining distance. You can use the last equation to calculate the maximum speed you could have to avoid a collision.
Use x = 39m and solve for v₀.
