Answer:
The object will travel at the speed of 16 m/s.
Explanation:
Given
To determine
How fast is the object traveling?
<u>Important Tip:</u>
The product of the mass and velocity of an object — momentum.
Using the formula
where
Thus, in order to determine the speed of the object, all we need to do is to substitute p = 64 and m = 4 in the formula
switch the equation
divide both sides by 4
simplify
m/s
Therefore, the object will travel at the speed of 16 m/s.
Is there any possible chance that at some point in your science
studies, sometime before you were given this question for your
homework, that maybe you might have encountered this formula
for the period of a simple pendulum ?
Period = (2 pi) √(length/gravity) .
If the length is 0.23 meter, and the
acceleration of gravity is 9.8 m/s²,
then the period is
= (2 pi) √(0.23/9.8)
= 0.963... second (rounded)
That's how long it takes for a simple pendulum, 23cm long,
hanging on a massless string and not swinging too far to
the side, to complete one full swing left and right.
Now, if you can figure out how many periods of 0.963 second
there are in 30 seconds, you'll have your answer. I'll leave
that part of it to you.
Answer:
Given that
V2/V1= 0.25
And we know that in adiabatic process
TV^န-1= constant
So
T1/T2=( V1 /V2)^ န-1
So = ( 1/0.25)^ 0.66= 2.5
Also PV^န= constant
So P1/P2= (V2/V1)^န
= (1/0.25)^1.66 = 9.98
A. RMS speed is
Vrms= √ 3RT/M
But this is also
Vrms 2/Vrms1= (√T2/T1)
Vrms2=√2.5= 1.6vrms1
B.
Lambda=V/4π√2πr²N
So
Lambda 2/lambda 1= V2/V1 = 0.25
So the mean free path can be inferred to be 0.25 times the first mean free path
C. Using
Eth= 3/2KT
So Eth2/Eth1= T2/T1
So
Eth2= 2.5Eth1
D.
Using CV= 3/2R
Cvf= Cvi
So molar specific heat constant does not change