Answer:
the force acting on the team mate is 1.19 kN.
Explanation:
given,
mass = 196 lbm
while tackling, the deceleration is from velocity 6.7 m/s to 0 m/s
time taken for deceleration = 0.5 sec
F = mass × acceleration
acceleration =
= -13.4 m/s²
1 lbs = 0.453 kg
196 lbs = 196 × 0.453 = 88.79 kg
F = 88.79 × 13.4
F = 1189.786 N = 1.19 kN
hence, the force acting on the team mate is 1.19 kN.
Answer:
1.65×10⁻⁶ N
Explanation:
m = Mass of ant =
r = Radius = 50 m
t = Time taken to complete on rotation = 60 seconds
Angular velocity
Centripetal acceleration is force that is acting outward
The magnitude of the upward force felt by the ant due to the second hand would be 1.65×10⁻⁶ N
Answer:
68.8 N 13.8°N of W
Explanation:
F₁ is 50 N 30°N of W. The terminal angle is 150°.
F₂ is 25 N 20°S of W. The terminal angle is -160°.
Graphically, you can add the vectors using head-to-tail method. Move F₂ so that the tail of the vector is at the head of F₁. The resultant vector will be from the tail of F₁ to the head of F₂.
Algebraically, find the x and y components of each vector.
F₁ₓ = 50 N cos(150°) = -43.3 N
F₁ᵧ = 50 N sin(150°) = 25 N
F₂ₓ = 25 N cos(-160°) = -23.5 N
F₂ᵧ = 25 N sin(-160°) = -8.6 N
The x and y components of the resultant vector are the sums:
Fₓ = -43.3 N + -23.5 N = -66.8 N
Fᵧ = 25 N + -8.6 N = 16.4 N
The magnitude of the resultant force is:
F = √(Fₓ² + Fᵧ²)
F = √((-66.8 N)² + (16.4 N)²)
F = 68.8 N
The direction of the resultant force is:
θ = tan⁻¹(Fᵧ / Fₓ)
θ = tan⁻¹(16.4 N / -66.8 N)
θ = 166.2°
θ = 13.8°N of W
