Refer to the figure shown below.
The velocity of the child and the velocity of the ship should be added vectorially to find the speed and direction of the child relative to the water surface.
The magnitude of the child's velocity is
v = √(2² + 18²) = 18.11 mph
The direction of the child's speed is
θ = tan⁻¹ (18/2) = tan⁻¹ 9 = 83.7° north of east or counterclockwise from the eastern direction.
Answer:
The magnitude is 18.1 mph.
The direction is 84° north of east.
Answer:
The acceleration of the both masses is 0.0244 m/s².
Explanation:
Given that,
Mass of one block = 602.0 g
Mass of other block = 717.0 g
Radius = 1.70 cm
Height = 60.6 cm
Time = 7.00 s
Suppose we find the magnitude of the acceleration of the 602.0-g block
We need to calculate the acceleration
Using equation of motion

Where, s = distance
t = time
a = acceleration
Put the value into the formula



Hence, The acceleration of the both masses is 0.0244 m/s².
Answer:
500 N
Explanation:
Natural bouyency will keep the diver in the same place, no matter what the mass of gfs. This means if the diver is displacing with a weight of 500 N, the upthrust will also be 500 N.