What is the frequency of a simple pendulum 2.7 m long in the following situations?
(a) in a room that is stationary hz
(b) in an elevator accelerating upward at a rate of 1.1 m/s2 hz
(c) in free fall hz

(a) 0.35Hz (b) 0.37Hz (c) 0

Explanation: frequency of a simple pendulum is expressed as ;

f= 1/2π√(a/L)

(a) in a room that is stationary, a, is the acceleration due to gravity =9.8m/s²

f= 1/2π√(g/L)

f= 1/2π( 9.8/2)

f= 0.35Hz

(b) If the pendulum is in an elevator accelerating at a rate of 1.1m/s²

The total force will surely to both a and g

Therefore apparent acceleration = a+g= 1.1 +9.8=10.9 m/s²

f= 1/2π √( 10.9/2)

f= 0.37Hz

(c) For a free fall, apparent acceleration is 0

f= 1/2π √ (0/L)

f = 0

get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm