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Paula has gone 1 km east as the total displacement. The total distance would be 7 km. Why? Well, the difference between displacement and distance is displacement is directional which is why it’s included in velocity and not speed. However, distance is more broad and not as specific.
Answer:
712.5 kg m/s
Explanation:
Work out the total momentum before the event (before the collision):
p = m × v
Massof Deon = 95kg
Velocity =7.5m/s
Mass of Chuck = 120kg
Velocity = 0m/s
Momentum of Deon before = 95 × 7.5 = 712.5 kg m/s
Momentum of Chuck before = 120 × 0 = 0 kg m/s
Total momentum before = 712.5+ 0 = 712.5 kg m/s
Working out the total momentum after the event (after the collision):
Because momentum is conserved, total momentum afterwards = 712.5 kg m/s
Answer:
- The process of comparing unknown quantities with known quantities is called measurement .
If it measurement system were not established ,then there may create lots of problem such as: unequal measure ,anyone can get more ,or anyone can get less mass ,so on .
Answer:
F_total = 10000 N
Explanation:
For this exercise we use that the force is a vector and the best way to do the sum is that since the two tugs pull the boat with the same intensity and angle, the sum of the component of the force perpendicular to the movement becomes zero.
The components parallel to the movement of the tugs is
∑ F = 2 Fcos θ
let's calculate
F_total = 2 10000 cos 60
F_total = 10000 N
Answer:
a) w = 4.24 rad / s
, b) α = 8.99 rad / s²
Explanation:
a) For this exercise we use the conservation of kinetic energy,
Initial. Vertical bar
Emo = U = m g h
Final. Just before touching the floor
Emf = K = ½ I w2
As there is no friction the mechanical energy is conserved
Emo = emf
mgh = ½ m w²
The moment of inertial of a point mass is
I = m L²
m g h = ½ (m L²) w²
w = √ 2gh / L²
The initial height h when the bar is vertical is equal to the length of the bar
h = L
w = √ 2g / L
Let's calculate
w = RA (2 9.8 / 1.09)
w = 4.24 rad / s
b) Let's use Newton's equation for rotational motion
τ = I α
F L = (m L²) α
The force applied is the weight of the object, which is at a distance L from the point of gro
mg L = m L² α
α = g / L
α = 9.8 / 1.09
α = 8.99 rad / s²
