Answer: 11 km/h at 339° compass
Explanation:
A sees B moving south at 0 km/h
A is moving north at 12cos30 = 10.392 km/h
Therefore B must be moving north at 10.392 k/h
A is moving east at 12sin30 = 6 km/h
B appears to be moving west at 10 km/h
Therefore B must be moving west at 10 - 6 = 4 km/h
B is moving v = √(4² + 10.392²) = 11.135... 11 km/h
θ = arctan( -4 / 10.392) = -21.05 = 339°
Answer:
Acceleration a ≤ 3 m/s^2
the greatest acceleration that the truck can have without losing its load is 3 m/s^2
Explanation:
For the truck to accelerate without losing its load.
Acceleration force of truck must be less than or equal to the maximum friction force between the truck bed and the load.
Fa ≤ F(friction)
But;
Fa = mass × acceleration
Fa = ma
ma ≤ F(friction)
a ≤ (F(friction))/m ......1
Given;
Fa = mass × acceleration
Fa = ma
mass m = 800 kg
F(friction) = 2400 N
Substituting the given values into equation 1;
a ≤ F(friction)/m
a ≤ 2400N/800kg
a ≤ 3 m/s^2
the greatest acceleration that the truck can have without losing its load is 3 m/s^2
A). It takes air in from outside the body.
