Answer:
i) E = 269 [MJ] ii)v = 116 [m/s]
Explanation:
This is a problem that encompasses the work and principle of energy conservation.
In this way, we establish the equation for the principle of conservation and energy.
i)
![W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]](https://tex.z-dn.net/?f=W_%7B1-2%7D%3D%20%28F%2Ad%29%20-%20%28m%2Ag%2Ah%29%5C%5CW_%7B1-2%7D%3D%28500000%2A2.5%2A10%5E3%29-%2840000%2A9.81%2A2.5%2A10%5E3%29%5C%5CW_%7B1-2%7D%3D%20269%2A10%5E6%5BJ%5D%20or%20269%20%5BMJ%5D)
At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.
Er = 269*10^6[J]
ii ) With the energy calculated at the previous point, we can calculate the speed developed.
![E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]](https://tex.z-dn.net/?f=E_%7Bk2%7D%3D0.5%2Am%2Av%5E2%5C%5C269%2A10%5E6%3D0.5%2A40000%2Av%5E2%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B269%2A10%5E6%7D%7B0.5%2A40000%7D%20%7D%5C%5C%20v%3D116%5Bm%2Fs%5D)
The solution for the problem is:
1 Watt = 1 Joule per second
1 Watt*second = 1 Joule
a Kilowatt is 1,000 Watts
an hour is 60 seconds times 60 minutes or 3,600 seconds
a Kilowatt * hour is 1,000 Watts in 3,600 seconds
15 W*h = 15,000 Watt*hour = 15,000 Watt * 3,600 seconds = 54,000,000
Watt*second
54,000,000 Watt*second = ? Joules
54,000,000 Joules / second = 54,000,000 Watts
Answer:
B
Explanation:
While answer C may sound correct, Answer B is makes more sense. We know you cant use High-beam lights when u cant see ongoing traffic because it could affect the other driver coming across from you. Its good to use it when legal and safe, but in that term I still don't believe there's no reason for HIGH-beamed. That's this leaves B, when you are on u lighted streets.
Answer:
a and b are the correct answers
Explanation:
