The correct answer would be 1.375 < t < 3 i hope this helps anyone
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:
270 m
Explanation:
Given:
v₀ = 63 m/s
a = 2.8 m/s²
t = 4.0 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (63 m/s) (4.0 s) + ½ (2.8 m/s²) (4.0 s)²
Δx = 274.4 m
Rounded to two significant figures, the displacement is 270 meters.
Answer:
2.2 s
Explanation:
Hi!
Let's consider the origin of the coordinate system at the ground, and consider that the clam starts with zero velocity, the equation of motion of the clam is given by
We are looking for a time t for which x(t) = 0
Solving for t:
Rounding at the first decimal:
t = 2.2 s
Answer:
Explanation:
There are two types of collision.
(a) Elastic collision: When there is no loss of energy during the collision, then the collision is said to be elastic collision.
In case of elastic collision, the momentum is conserved, the kinetic energy is conserved and all the forces are conservative in nature.
The momentum of the system before collision = the momentum of system after collision
The kinetic energy of the system before collision = the kinetic energy after the collision
(b) Inelastic collision: When there is some loss of energy during the collision, then the collision is said to be inelastic collision.
In case of inelastic collision, the momentum is conserved, the kinetic energy is not conserved, the total mechanical energy is conserved and all the forces or some of the forces are non conservative in nature.
The momentum of the system before collision = the momentum of system after collision
The total mechanical energy of the system before collision = total mechanical of the system after the collision
