Answer:
The angular frequency of the block is ω = 5.64 rad/s
Explanation:
The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.
Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.
The angular frequency of the oscillation ω is
ω = v/r
ω = 62 cm/s ÷ 11 cm
ω = 5.64 rad/s
So, the angular frequency of the block is ω = 5.64 rad/s
Measuring density: Measure the mass (in grams) of each mineral sample available to you. The mass of each sample is measured using a balance or electronic scale. Record mass on a chart.
It would be 12hz because it
Answer:
a = 0.55 m / s²
Explanation:
The centripetal acceleration is given by the relation
a = v² / r
angular and linear velocities are related
v = w r
we substitute
a = w² r
In the exercise they indicate the angular velocity w = 1 rev/min, let's reduce to the SI system
w = 1 rev / min (2pi rad / 1rev) (1min / 60s) = 0.105 rad/ s
let's calculate
a = 0.105² 50.0
a = 0.55 m / s²
I believe the correct answer would be kimberlite. Diamonds are usually found in pipes 50 to 200 m across made of kimberlite. It is an igneous rock that is known to contain traces of diamonds. It is named base on the town where it was discovered which is Kimberley, South Africa.
