Answer:
0.34 sec
Explanation:
Low point of spring ( length of stretched spring ) = 5.8 cm
midpoint of spring = 5.8 / 2 = 2.9 cm
Determine the oscillation period
at equilibrum condition
Kx = Mg
g= 9.8 m/s^2
x = 2.9 * 10^-2 m
k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93
note : w = 
= 
= 18.38 rad/sec
Period of oscillation = 
= 0.34 sec
Answer:
Scientists plan to release a space probe that will enter the atmosphere of a gaseous planet. The temperature of the gaseous planet increases linearly with the height of the atmosphere as measured from the top of a visible boundary layer, defined as 0 kilometers in altitude. The instruments on board can withstand a temperature of 601 K. At what altitude will the probe's instruments fail? A. 50 kilometers B. 80 kilometers C. 83 kilometers D. 100 kilometers E. 111 kilometers
Explanation:
A. 50 kilometers
maximum speed of cheetah is
speed of gazelle is given as
Now the relative speed of Cheetah with respect to Gazelle
now the relative distance between Cheetah and Gazelle is given initially as "d"
now the time taken by Cheetah to catch the Gazelle is given as
so by rearranging the terms we can say
so above is the relation between all given variable
Answer:
Remain the same
Explanation:
There is no relationship between amplitude frequency.
Answer:
18kg dog walking
Explanation:
the dog walking has more inertia than the boy and girl because its gravity is pulling it down to the earth
