Answer:
The change in energy of the gas during the process is 
joules.
Explanation:
We can represent this process by the First Law of Thermodynamics, in which gas does work on its surroundings and absorbs heat from there to describe its change in energy. In other words:
Where:
- Heat absorbed by the gas, measured in joules.
- Work done by the gas, measured in joules.
- Change in energy, measured in joules.
If we know that 
and 
, the change in energy of the gas is:
The change in energy of the gas during the process is joules.
Answer:
pascal
Explanation:
its obtained after either division or multiplication
Point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.
On the graph, A is the point where magnitude of the acceleration of the particle is greatest as compared to other positions on the graph because the height of point A is the largest as compared to other points of the graph.
The graph shows at which point acceleration of an object is higher and lower so we can conclude that point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.
Learn more about acceleration here: brainly.com/question/933224
Learn more: brainly.com/question/25887663
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)
Answer:
Explanation:
a ) V = 3 cos(0.5t)
differentiating with respect to t
dv /dt = -3 x .5 sin0.5t
= -1.5 sin0.5t.
acceleration = - 1.5 sin 0.5t
when t = 3 s
acceleration = - 1.5 sin 1.5
= - 1.496 ms⁻²
v = 3 cos.5t
b ) dx/dt = 3 cos 0.5 t
dx = 3 cos 0.5 t dt
integrating on both sides
x = 3 sin .5t / .5
x = 6 sin0.5t
At t = 2 s
x = 6 sin 1
x = 5.05 m
