Explanation:
Given:
v₀ = 0 m/s
v = 49 m/s
a = 9.8 m/s²
Find: t
v = at + v₀
49 m/s = (9.8 m/s²) t + 0 m/s
t = 5 s
When naming isotopes, the number beside it indicates the mass number. Now mass number is the sum of protons and neutrons.
We know that the mass number of a stable Sulfur atom is 32, so we can rule out C as an isotope. The atomic number of Sulfur is 16 and that means that there are 16 protons. Now if we subtract 16 from 33, then that means there are 17 neutrons. So we can rule out D.
Phosphorus on the other hand has an atomic number of 15. If we subtract 15 from 31 we will have 16. That means that Phosphorus-31 has 16 neutrons.
The answer would then be A. Phosphorus 31.
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:
Its rotation will be 3.89x10⁴ rad/s.
Explanation:
We can find the rotation speed by conservation of the angular momentum:

(1)
The initial angular speed is:

The moment of inertia (I) of a sphere is:
(2)
Where m is 9 times the sun's mass and r is the sun's radius
By entering equation (2) into (1) we have:

Hence, its rotation will be 3.89x10⁴ rad/s.
I hope it helps you!