D. A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she hangs it from a scale and finds its weig
1 answer:
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Complete question:
A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find
(a) the force constant of the spring and (b) the amplitude of the motion.
Answer:
(a) the force constant of the spring = 47 N/m
(b) the amplitude of the motion = 0.292 m
Explanation:
Given;
mass of the spring, m = 200g = 0.2 kg
period of oscillation, T = 0.410 s
total mechanical energy of the spring, E = 2 J
The angular speed is calculated as follows;
(a) the force constant of the spring
(b) the amplitude of the motion
E = ¹/₂kA²
2E = kA²
A² = 2E/k
Answer:
The weight of measuring stick is 9.8 N
Explanation:
given information:
the mass of the rock, = 1 kg
measuring stick, x =1 m
d = 0.25 m
to find the weight of measuring stick, we can use the following equation:
τ = Fd
τ = 0
-
= 0
F_{r} = the force of the rock
F_{s} = the force of measuring stick
= m g
= 1 kg x 9.8 m/s
= 9.8 N
thus, the weight of measuring stick is 9.8 N
Answer:
1.24611
Explanation:
V = Velocity = 10 ft/s
L = Length = 2 ft
g = Acceleration due to gravity = 32.2 ft/s²
Froude number is given by
Converting to SI units
The Froude number is 1.24611
The Froude number is equal. The Froude number is dimensionless as the units cancel each other. In order for this to happen the units used need to be consitent either imperial or SI.