Answer:
taking a shower brushing your teeth and washing your hands
Explanation:
Answer:
F = 1.24*10^4 N
Explanation:
Given
Depth of the ship, h = 25 m
Density of water, ρ = 1.03*10^3 kg/m³
Diameter of the hatch, d = 0.25 m
Pressure of air, P(air) = 1 atm
Pressure of water =
P(w) = ρgh
P(w) = 1.03*10^3 * 9.8 * 25
P(w) = 2.52*10^5 N/m²
P(net) = P(w) + P(air) - P(air)
P(net) = P(w)
P(net) = 2.52*10^5 N/m²
Remember,
Pressure = Force / Area, so
Force = Area * Pressure
Area = πr² = πd²/4
Area = 3.142 * 0.25²/4
Area = 3.142 * 0.015625
Area = 0.0491 m²
Force = 0.0491 * 2.52*10^5
F = 12373 N
F = 1.24*10^4 N
Answer:
The farther star will appear 4 times fainter than the star that is near to the observer.
Explanation:
Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time
Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)
This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by
For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by
Hence we sense it as 4 times fainter than the nearer star.
Ok so here is the thing. It is necessary to introduce the atomic number Z into the following equation and the reason for that is that we are not working here with hydrogen (H). It will go like this:
<span>E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )| </span>
<span>E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 J </span>
<span>After that we need to plug the E value calculated into the equation. Remember that the wavelength is always positive:</span>
<span>E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 m </span>
so 6.56×10^-7 m or better written 656 nm is in the visible spectrum
