The answer should be flammability
Answer:
a = 1.1 m/s^2
Explanation:
let a be the acceleration of the astronaut, M be the mass of the earth and r be the radius of the earth.
from the information provide:
the distance r is the radius of the earth.
the distance ,d, the astronaut is from the surface of the earth and the only force acting on the astronaut is gravity, the astronaut accelerates with due to gravity only that is d = 3×r because we consider the earth center as the center of mass and gravity attracts object towards the center of mass.
then, we know that:
a = GM/d^2
= (6.67×10^-11)(5.972×10^24)/[(3×6371×10^3)^2]
= 1.1 m/s^2
Therefore, the Astronaut will be accelerating with an acceleration of 1.1 m/s^2 towards the earth.
Given: Mass of earth Me = 5.98 x 10²⁴ Kg
Radius of earth r = 6.37 x 10⁶ m
G = 6.67 x 10⁻¹¹ N.m²/Kg²
Required: Smallest possible period T = ?
Formula: F = ma; F = GMeMsat/r² Centripetal acceleration ac = V²/r
but V = 2πr/T
equate T from all equation.
F = ma
GMeMsat/r² = Msat4π²/rT²
GMe = 4π²r³/T²
T² = 4π²r³/GMe
T² = 39.48(6.37 x 10⁶ m)³/6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)
T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²
T² = 25,563,909.77 s²
T = 5,056.08 seconds or around 1.4 Hour
There is a covalent bond formed between two atoms that do not easily lose electrons.
hope this helps :)
Answer: 3MeV electron
Explanation:
m_e={9.1\times 10^{-31} m_α=4\times m_e m_a={9.1\times 10^{-31}
m_p=1.67\times 10^{-27}
(a) K.E. Energy of electron =
=3MeV
=1.05\times10^{18}
(b) K.E. Energy of alpha particle =
=10MeV
=0.88\times10^{18}
(c) K.E. Energy of auger particle =
=0.1MeV
=0.035\times10^{18}
(d) K.E. Energy of proton particle =
=400keV
=0.766\times10^{14}
![v_p=\sqrt{0.766\times10^{14} } =0.875\times10^{7}[tex]\frac{m}{s}](https://tex.z-dn.net/?f=v_p%3D%5Csqrt%7B0.766%5Ctimes10%5E%7B14%7D%20%7D%20%3D0.875%5Ctimes10%5E%7B7%7D%5Btex%5D%5Cfrac%7Bm%7D%7Bs%7D)
from (a),(b),(c),and (d) we can clearly say that the velocity of the electron is more so the penetration of the electron will be deepest.
