**Answer:**

The weight of the probe is 50 Newtons

**Explanation:**

Newtons second law states that F = ma

Given the mass of 25kg, and the **a**cceleration of 2m/s^2, we can substitute both values into the equation to find the weight **f**orce.

The weight of the probe is 50 Newtons

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**Answer:**

a) F = 1.26 10⁵ N, b) F = 2.44 10³ N, c) F_net = 1.82 10³ N directed vertically upwards

**Explanation:**

For this exercise we must use the relationship between momentum and momentum

I = Δp

F t = p_f -p₀

a) It asks to find the force

as the man stops the final velocity is zero

F = 0 - p₀ / t

the speed is directed downwards which is why it is negative, therefore the result is positive

F = m v₀ / t

F = 63.5 7.89 / 3.99 10⁻³

F = 1.26 10⁵ N

b) in this case flex the knees giving a time of t = 0.205 s

F = 63.5 7.89 / 0.205

F = 2.44 10³ N

c) The net force is

F_net = Sum F

F_net = F - W

F_net = F - mg

let's calculate

F_net = 2.44 10³ - 63.5 9.8

F_net = 1.82 10³ N

since it is positive it is directed vertically upwards