Answer:
Part a)
T = 0.52 s
Part b)
Part c)
Explanation:
As we know that the particle move from its maximum displacement to its mean position in t = 0.13 s
so total time period of the particle is given as
now we have
Part a)
T = time to complete one oscillation
so here it will move to and fro for one complete oscillation
so T = 0.52 s
Part b)
As we know that frequency and time period related to each other as
Part c)
As we know that
wavelength = 1.9 m
frequency = 1.92 Hz
so wave speed is given as
Answer:
In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles, so an object will weigh approximately 0.5% more at the poles than at the Equator.
Complete question:
A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?
Answer:
the weight of the air is 76.44 lbs
Explanation:
Given;
dimension of the dormitory, = 14 ft by 13 ft by 6 ft
density of the air, = 0.07 lbs/ft³
The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft
= 1092 ft³
The weight of the air = density x volume
= 0.07 lbs/ft³ x 1092 ft³
= 76.44 lbs
Therefore, the weight of the air is 76.44 lbs
V^2= 2* g * s
V^2 = 2 * 9.8 * 50 = 980
v = square root (980)=...... m/s
Answer:
Initial position of a body is the position of the body before accelerating or increasing its velocity the position changes and then that position is the final position.
hope it is helpful...
