Answer:
<em> The object has frequency of 2 Hz and time period of 0.5 s.</em>
Explanation:
<em>Frequency</em> is defined as number of oscillation per second ie back and forth swings done in single second.
Here it is given that the object oscillates 20 times in 10 seconds.
So f =
= 2Hz
The <em>time period</em> is defined as time taken by the object to complete one full oscillation.
T = 
T=
=0.5 s
<em>Thus the object has frequency of 2 Hz and time period of 0.5 s.</em>
We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"
- it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
From the question we are told
the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?
Generally the equation for the Force is mathematically given as
F=\frac{F}{dx}
Therefore
F=-kdx
k=600Nm^{-1}
now
K.E=0.5x ds^2
K.E=600*(-0.1^2)
K.E=3J
Therefore
the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
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Answer:
a) 
b) 
c) 
Explanation:
From the question we are told that
Distance to Betelgeuse 
Mass of Rocket 
Total Time in years traveled 
Total energy used by the United States in the year 2000 
Generally the equation of speed of rocket v mathematically given by


where




Therefore


b)
Generally the equation of the energy E required to attain prior speed mathematically given by


c)Generally the equation of the energy
required to accelerate the rocket mathematically given by



The speed of car is 100.8km/h









v car= 28×3.6
=100.8km/h
Hence, the speed of the car is 100.8km/h
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Answer:
umm the lower the frequency the higher the pitch
Explanation: