It would be C)Whales learn their migration paths from other whales
A) the forces are acting in the same direction.. B) Together, forces are acting in opposite directions
Answer:
A) 80 N
B) 20 N
Explanation:
A) If the forces acting are in the same direction, then the net force will be a sum of both so many faces..
Thus;
ΣF = 50 + 30
ΣF = 80 N
B) If the forces are acting in the in opposite directions with the larger force pointing in the positive y-axis then, the net force is;
ΣF = 50 - 30
ΣF = 20 N
Acceleration means speeding up, slowing down, or changing direction. The graph doesn't show anything about direction, so we just have to examine it for speeding up or slowing down ... any change of speed.
The y-axis of this graph IS speed. So the height of a point on the line is speed. If the line is going up or down, then speed is changing.
Sections a, c, and d are all going up or down. Section b is the only one where speed is not changing. So we can't be sure about b, because we don't know if the track may be curving ... the graph can't tell us that. But a, c, and d are DEFINITELY showing acceleration.
His. Curbs I b h bs. H b u b
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so 
= A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
