Answer:
161.86 N
Explanation:
mass of box m= 55.0 kg
weight of the box, mg= 55×9.81
g here is acceleration due to gravity =9.81 m/sec^2
coefficient of friction between the box and the surface μ= 0.3
the friction force F_s= μmg= 0.3×55×9.81
=161.86 N
to move the ball horizontal force required is 161.86 N
Answer:
a) a = 3.06 10¹⁵ m / s
, b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N
Explanation:
This exercise will average solve using the moment relationship.
a ) let's use the relationship between momentum and momentum
I = ∫ F dt = Δp
F t = m
- m v₀
F = m (v_{f} -v₀o) / t
in the exercise indicates that the speed module is the same, but in the opposite direction
F = m (-2v) / t
if we use Newton's second law
F = m a
we substitute
- 2 mv / t = m a
a = - 2 v / t
let's calculate
a = - 2 4.59 10²/3 10⁻¹³
a = 3.06 10¹⁵ m / s
b) F= m a
F= 4.68 10⁻²⁶ 3.06 10¹⁵
F= 1.43 10⁻¹⁰ N
c) if we hit the wall for 1015 each exerts a force F
F_total = n F
F_total = n m a
F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵
F_total = 14.32 10⁻²⁶ N
Answer:
the force exerted by the seat on the pilot is 10766.7 N
Explanation:
The computation of the force exerted by the seat on the pilot is as follows:

Hence, the force exerted by the seat on the pilot is 10766.7 N
Answer:
≈ 20.35 N [newton's of tension]
Explanation:
( (2.9 × 9.8) ÷ cos(35.6°) ) × sin (35.6°) =
( (28.42) ÷ (≈0.813) ) × (≈0.582) =
(≈34.96) × (≈0.582) = 20.3449446.... ≈ 20.35