The hawk’s centripetal acceleration is 2.23 m/s²
The magnitude of the acceleration under new conditions is 2.316 m/s²
radius of the horizontal arc = 10.3 m
the initial constant speed = 4.8 m/s
we know that the centripetal acceleration is given by
= 
= 23.04/10.3
= 2.23 m/s²
It continues to fly but now with some tangential acceleration
= 0.63 m/s²
therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration
so
= 
= 
= 2.316 m/s²
So the magnitude of net acceleration will become 2.316 m/s².
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<u>Answer</u>
3.7 Km south
<u>Explanation</u>
The definition of displacement is the distance traveled in a specific direction. It is the vector quantity. We add displacements like the way we add vectors.
Taking the direction towards North to be positive (+1.7 Km), the distance towards south would be negative (-5.4 Km).
Now lets add the two values.
(+1.7) + (-5.4) = 1.7 - 5.4
= - 3.7 Km But negative was towards south.
∴ Answer = 3.7 Km south.
Answer:
θ = 4.78º
with respect to the vertical or 4.78 to the east - north
Explanation:
This is a velocity compound exercise since it is a vector quantity.
The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed
v_fly² = v_nort² + v_air²
v_nort² = v_fly² + - v_air²
Let's use trigonometry to find the direction of the plane
sin θ = v_air / v_fly
θ = sin⁻¹ (v_air / v_fly)
let's calculate
θ = sin⁻¹ (10/120)
θ = 4.78º
with respect to the vertical or 4.78 to the north-east
Pretty sure it’s false....................
Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.
Explanation:
Given: Mass = 5 kg
Spring constant = 6 N/m
Formula used to calculate period is as follows.

where,
T = period
m = mass
k = spring constant
Substitute the values into above formula as follows.

Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.