B. all of nature follows uniform, unchaining laws.
Answer:
The answer to your question is given below
Explanation:
Since both object A and B were dropped from the same height and the air resistance is negligible, both object A and B will get to the ground at the same time.
From the question, we were told that object A falls through a distance to dA at time t and object B falls through a distance of dB at time 2t.
Remember, both objects must get to the ground at the same time..!
Let the time taken for both objects to get to the ground be t.
Time A = Time B = t
But B falls through time 2t
Therefore,
Time A = Time B = 2t
Height = 1/2gt^2
For A:
Time = 2t
dA = 1/2 x g x (2t)^2
dA = 1/2g x 4t^2
For B
Time = t
dB = 1/2 x g x t^2
Equating dA and dB
dA = dB
1/2g x 4t^2 = 1/2 x g x t^2
Cancel out 1/2, g and t^2
4 = 1
4dA = dB
Divide both side by 4
dA = 1/4 dB
Answer:
This question is incomplete
Explanation:
This question is incomplete because of the absence of the time taken to complete one full cycle.
Frequency (<em>f</em>) will be calculated first as
<em>f </em>= <em>N </em>÷<em> t</em>
where <em>N </em>is the number of cycles and <em>t </em>is the time taken to complete one full cycle. The unit for frequency is Hertz (Hz).
To calculate the period, <em>T, </em>the formula below will be used
<em>T </em>= 1 ÷ <em>f</em>
The unit for period is secs
Answer:
7.65 m
Explanation:
= Initial pressure = 0.03 atm
= Final pressure = 1 atm
= Inital radius = 21 m
= Intial volume of gas = 
= Final volume of gas = 
= Initial temperature = 200 K
= Final temperature = 323 K
From ideal gas law we have
The radius at liftoff is 7.65 m
