Answer: The car has a kinetic energy (because it's in motion) of: 
Explanation:
Answer:
Acceleration = 0.0282 m/s^2
Distance = 13.98 * 10^12 m
Explanation:
we will apply the energy theorem
work done = ΔK.E ( change in Kinetic energy ) ---- ( 1 )
<em>where :</em>
work done = p * t
= 15 * 10^6 watts * ( 1 year ) = 473040000 * 10^6 J
( note : convert 1 year to seconds )
and ΔK.E = 1/2 mVf^2 given ; m = 1200 kg and initial V = 0
<u>back to equation 1 </u>
473040000 * 10^6 = 1/2 mv^2
Vf^2 = 2(473040000 * 10^6 ) / 1200
∴ Vf = 887918.92 m/s
<u>i) Determine how fast the rocket is ( acceleration of the rocket )</u>
a = Vf / t
= 887918.92 / ( 1 year )
= 0.0282 m/s^2
<u>ii) determine distance travelled by rocket </u>
Vf^2 - Vi^2 = 2as
Vi = 0
hence ; Vf^2 = 2as
s ( distance ) = Vf^2 / ( 2a )
= ( 887918.92 )^2 / ( 2 * 0.0282 )
= 13.98 * 10^12 m
Answer:
2.61 atm
Ley de Boyle
Explanation:
= Presión inicial = 0.96 atm
= Presión final
= Volumen inicial = 95 mL
= Volumen final = 35 mL
En este problema usaremos la ley de Boyle.
La presión ejercida sobre el émbolo para reducir su volumen es de 2.61 atm.
Answer:
The speed it reaches the bottom is
Explanation:
Given: 
, 
Using the conservation of energy theorem
, 
![m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2](https://tex.z-dn.net/?f=m%2Ag%2Ah%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2A%28r%2Aw%29%5E2%20%2B%5Cfrac%7B1%7D%7B2%7D%2A%5B%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Ar%5E2%5D%2Aw%5E2)
Solve to w'
TRUE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
