Answer:
VAB = 20km/hr
Explanation:
<u>Given the following data;</u>
Velocity of car A, VA = 60km/hr
Velocity of car B, VB = 80km/hr
To find the relative velocity of B w.r.t A, VAB;
Since the two cars are moving in the same direction, we have;
VAB = VB - VA
Substituting into the equation, we have;
VAB = 80 - 60
<em>VAB = 20km/hr</em>
Therefore, the relative velocity of car B with respect to car A is 20 kilometers per hour.
Here it is an application of Newton's III law
as we know by Newton's III law that every action has equal and opposite reaction
So here as we know that two boys jumps off the boat with different forces
from front side of the boat the boy jumps off with force 45 N which means as per Newton's III law if boy has a force of 45 N in forward direction then he must apply a reaction force on the boat in reverse direction of same magnitude
So boat must have an opposite force on front end with magnitude 45 N
Now similar way we can say
from back side of the boat the boy jumps off with force 60 N which means as per Newton's III law if boy has a force of 60 N in backward direction then he must apply a reaction force on the boat in reverse direction of same magnitude
So boat must have an opposite force on front end with magnitude 60 N
So here net force due to both jump on the boat is given by



so boat will have net force F = 15 N in forward direction due to both jumps
Answer:
a) 37.70 m/s
b)710.6 m/s²
Explanation:
Given that ;
Mass of object = 2 kg
Radius of the motion = 2m
Frequency of motion = 3 rev/s
The formula to apply is;
v= 2πrf where v is linear speed
v = 2×π×2×3 =12π = 37.70 m/s
Centripetal acceleration is given as;
a= 4×π²×r×f²
a= 4×π²×2×3²
a=710.6 m/s²
south = -(north)
Displacement = (4 km north) + (2 km south) + (5 km north) + (5 km south)
Displacement = (4 km north) - (2 km north) + (5 km north) - (5 km north)
Displacement = (4 - 2 + 5 - 5) km north
<u>Displacement = 2 km north </u>