Complete question :
NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supply storage area of the lunar outpost where gravity is 1.63m/s/s can only support 1 x 10 over 5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost?
Answer:
601000 N
Explanation:
Given that :
Acceleration due to gravity at lunar outpost = 1.6m/s²
Supported Weight of supplies = 1 * 10^5 N
Acceleration due to gravity on the earth surface = 9.8m/s²
Maximum weight of supplies as measured on EARTH :
Ratio of earth gravity to lunar post gravity:
(Earth gravity / Lunar post gravity) ;
(9.8 / 1.63) = 6.01
Hence, maximum weight of supplies as measured on EARTH should be :
6.01 * (1 × 10^5)
6.01 × 10^5
= 601000 N
Answer:
None
Explanation:
Subatomic particles are the particles which are very smaller than the atoms. Elementary particles are the examples of subatomic particles.
Elementary particles are the particles without any sub-structure which means they are not composed of other particles.
The elementary particles are classified into three categories which are discussed below:
(1) Quarks: up, down, top, bottom, strange, and charm.
(2) Leptons: muon, muon neutrino, electrons, electron neutrino, tau, tau neutrino.
(3) Bosons: Z bosons, W bosons, Higgs, Gluon, photons.
Mesons are the particles which compose one quark and one anti quarks.
Therefore, in the given list there is no meson.
Answer:
248
Explanation:
L = Inductance of the slinky = 130 μH = 130 x 10⁻⁶ H
= length of the slinky = 3 m
N = number of turns in the slinky
r = radius of slinky = 4 cm = 0.04 m
Area of slinky is given as
A = πr²
A = (3.14) (0.04)²
A = 0.005024 m²
Inductance is given as
N = 248
Answer:
a. k = (1/k₁ + 1/k₂)⁻¹ b. k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹
Explanation:
Since only one force F acts, the force on spring with spring constant k₁ is F = k₁x₁ where x₁ is its extension
the force on spring with spring constant k₂ is F = k₂x₂ where x₁ is its extension
Let F = kx be the force on the equivalent spring with spring constant k and extension x.
The total extension , x = x₁ + x₂
x = F/k = F/k₁ + F/k₂
1/k = 1/k₁ + 1/k₂
k = (1/k₁ + 1/k₂)⁻¹
B
The force on spring with spring constant k₃ is F = k₃x₃ where x₃ is its extension
Let F = kx be the force on the equivalent spring with spring constant k and extension x.
The total extension , x = x₁ + x₂ + x₃
x = F/k = F/k₁ + F/k₂ + F/k₃
1/k = 1/k₁ + 1/k₂ + 1/k₃
k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹
