In order to give the apple more kinetic energy than it has right now,
you must make it move faster than it is moving right now. You could
drop it, roll it, toss it, or gingerly place it in a moving car or airplane.
Whatever method you select, motion ... specifically speed ... is the
key to kinetic energy.
Answer:
The molecules in hot air are moving faster than the molecules in cold air.
Answer:
29.38 seconds
Explanation:
Half life, T = 22.07 s
No = 1293
Let N be the number of atoms left after time t
N = 1293 - 779 = 514
By the use of law of radioactivity
![N=N_{0}e^{-\lambda t}](https://tex.z-dn.net/?f=N%3DN_%7B0%7De%5E%7B-%5Clambda%20t%7D)
Where, λ is the decay constant
λ = 0.6931 / T = 0.6931 / 22.07 = 0.0314 decay per second
so,
![514=1293e^{-0.0314t}](https://tex.z-dn.net/?f=514%3D1293e%5E%7B-0.0314t%7D)
![2.5155=e^{0.0314t}](https://tex.z-dn.net/?f=2.5155%3De%5E%7B0.0314t%7D)
take natural log on both the sides
0.9225 = 0.0314 t
t = 29.38 seconds
Answer:
a ) 2.368 rad/s
b) 3.617 rad/s
Explanation:
the minimum angular velocity that Prof. Stefanovic needs to spin the bucket for the water not to fall out can be determined by applying force equation in a circular path
i.e
------ equation (1)
where;
![F_{inward} = m *a_c](https://tex.z-dn.net/?f=F_%7Binward%7D%20%3D%20m%20%2Aa_c)
![F_{inward} = m*r* \omega^2](https://tex.z-dn.net/?f=F_%7Binward%7D%20%3D%20m%2Ar%2A%20%5Comega%5E2)
Also
![F_G = m*g](https://tex.z-dn.net/?f=F_G%20%3D%20m%2Ag)
since; that is the initial minimum angular velocity to keep the water in the bucket
Now; we can rewrite our equation as :
![mr \omega^2= mg + 0\\\omega^2 = \frac{m*g}{m*r}\\\omega^2 = \frac{g}{r}\\\omega = \sqrt{\frac{g}{r} \ \ } ------ equation \ \ \ {2}](https://tex.z-dn.net/?f=mr%20%5Comega%5E2%3D%20mg%20%2B%200%5C%5C%5Comega%5E2%20%3D%20%5Cfrac%7Bm%2Ag%7D%7Bm%2Ar%7D%5C%5C%5Comega%5E2%20%3D%20%5Cfrac%7Bg%7D%7Br%7D%5C%5C%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7Bg%7D%7Br%7D%20%5C%20%5C%20%20%7D%20%20%20%20%20------%20equation%20%5C%20%5C%20%5C%20%7B2%7D)
So; Given that:
The rope that is attached to the bucket is lm long and his arm is 75 cm long.
we have our radius r = 1 m + 75 cm
= ( 1 + 0.75 ) m
= 1.75 m
g = acceleration due to gravity = 9.81 m/s²
Replacing our values into equation (2) ; we have:
![\omega = \sqrt{ \frac{9.81}{1.75}}\\\omega = 2.368 \ rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%20%5Cfrac%7B9.81%7D%7B1.75%7D%7D%5C%5C%5Comega%20%3D%202.368%20%5C%20%20rad%2Fs)
b) if he detaches the rope and spins the bucket by holding it with his hand ; then the radius = 0.75 m
∴
![\omega = \sqrt{ \frac{9.81}{0.75}}\\\omega = 3.617 \ rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%20%5Cfrac%7B9.81%7D%7B0.75%7D%7D%5C%5C%5Comega%20%3D%203.617%20%5C%20%20rad%2Fs)