A simple pendulum consists of a point mass suspended by a weightless, rigid wire in a uniform gravitation field. Check all that
1 answer:
Answer:
f.The period is independent of the suspended mass.
Explanation:
The period of a pendulum is given by
where
L is the length of the pendulum
g is the acceleration due to gravity
From the formula, we see that:
1) the period of the pendulum depends only on its length, L, and it is proportional to the square root of the length
2) the period does not depend neither on the mass of the pendulum, nor on its amplitude of oscillation
So, the only correct statements are
f.The period is independent of the suspended mass.
Note: statement "e.The period is proportional to the length of the wire" is also wrong, because the period is NOT proportional to the length of the wire, but it is proportional to the square root of it.
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m = mass of the box
N = normal force on the box
f = kinetic frictional force on the box
a = acceleration of the box
μ = coefficient of kinetic friction
perpendicular to incline , force equation is given as
N = mg Cos30 eq-1
kinetic frictional force is given as
f = μ N
using eq-1
f = μ mg Cos30
parallel to incline , force equation is given as
mg Sin30 - f = ma
mg Sin30 - μ mg Cos30 = ma
"m" cancel out
a = g Sin30 - μ g Cos30
inserting the values
1.20 = (9.8) Sin30 - (9.8) Cos30 μ
μ = 0.44
The area of the circle with radius r is
A = πr²
The rate of change of area with respect to time is
The rate of change of the radius is given as
Therefore
When r = 10 ft, obtain
Answer: - 40π ft²/s (or - 127.5 ft²/s)
A = πr²
The rate of change of area with respect to time is
The rate of change of the radius is given as
Therefore
When r = 10 ft, obtain
Answer: - 40π ft²/s (or - 127.5 ft²/s)