Answer:
period of oscillations is 0.695 second
Explanation:
given data
mass m = 0.350 kg
spring stretches x = 12 cm = 0.12 m
to find out
period of oscillations
solution
we know here that force
force = k × x .........1
so force = mg = 0.35 (9.8) = 3.43 N
3.43 = k × 0.12
k = 28.58 N/m
so period of oscillations is
period of oscillations = 2π × ................2
put here value
period of oscillations = 2π ×
period of oscillations = 0.6953
so period of oscillations is 0.695 second
A. 1/9
Explanation:
The gravitational force between two objects is given by
where
G is the gravitational constant
m1 and m2 are the two masses
r is the distance between the two masses
From the formula, we see that the magnitude of the force is inversely proportional to the square of the distance: therefore, if the distance is tripled (increased by a factor 3), the magnitude of the force changes by a factor
The butterfly takes a vertical perpendicular path equivalent to 9m and travels a horizontal distance of 17m. The net path between the two is equivalent to that of the hypotenuse, so we will apply the Pythagorean theorem.
Therefore the magnitude of the butterfly's displacement is 9m
