Answer:
492.6 kN
Explanation:
Since the window as a radius of 2 m, its diameter is thus 4 m. Since the diving pool is 8 m deep, the height of water from the top of the window to the bottom of the pool is 8 - 4 = 4 m. The actual pressure acting on the window is thus
P = ρgh were ρ = density of water = 1000 kg/m³ g = 9.8 m/s² and h = 4 m
P = 1000 kg/m³ × 9.8 m/s² × 4 m = 39200 N/m².
Since P = F/A were F = force and A = area,
F = PA were A = area of window = πr² = π2² = 4π m² = 12.57 m²
F = 39200 N/m² × 12.57 m² = 492601.73 N = 492.6 kN
Hi there!
![\large\boxed{10 m/s}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B10%20m%2Fs%7D%7D)
When a projectile is launched, the HORIZONTAL component remains constant.
The force of gravity only changes the vertical component of the velocity.
As the two components are completely independent of one another, the horizontal component is UNCHANGED.
Answer:
The answer is "
"
Explanation:
Z=2, so the equation is ![E= \frac{-4B}{n^2}](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B-4B%7D%7Bn%5E2%7D)
Calculate the value for E when:
n=2 and n=9
The energy is the difference in transformation, name the energy delta E Deduct these two energies
In this transition, the wavelength of the photon emitted is:
![\Delta E=2.18 \times 10^{-18} ( \frac{1}{4}- \frac{1}{81})](https://tex.z-dn.net/?f=%5CDelta%20E%3D2.18%20%5Ctimes%20%2010%5E%7B-18%7D%20%28%20%5Cfrac%7B1%7D%7B4%7D-%20%5Cfrac%7B1%7D%7B81%7D%29)
![\lambda = \frac{h c}{\Delta E}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bh%20c%7D%7B%5CDelta%20E%7D)
![h ( Planck's\ constant) = 6.62 \times 10^{(-34)} \ Js \\\\ speed \ of \ light = 3 \times 10^{8} \ \frac{m}{s}\\\\= \frac{6.62 \times 10^{(-34)} \times 3 \times 10^ {8}}{2.18 \times 10^{-18}} (\frac{1}{4}- \frac{1}{81}) \\\\=3.83 \times 10^9 \ m\\\\](https://tex.z-dn.net/?f=h%20%28%20Planck%27s%5C%20%20constant%29%20%3D%206.62%20%5Ctimes%20%2010%5E%7B%28-34%29%7D%20%5C%20Js%20%5C%5C%5C%5C%20speed%20%5C%20of%20%5C%20light%20%3D%203%20%5Ctimes%2010%5E%7B8%7D%20%5C%20%5Cfrac%7Bm%7D%7Bs%7D%5C%5C%5C%5C%3D%20%5Cfrac%7B6.62%20%5Ctimes%2010%5E%7B%28-34%29%7D%20%5Ctimes%203%20%5Ctimes%2010%5E%20%7B8%7D%7D%7B2.18%20%5Ctimes%20%2010%5E%7B-18%7D%7D%20%28%5Cfrac%7B1%7D%7B4%7D-%20%5Cfrac%7B1%7D%7B81%7D%29%20%5C%5C%5C%5C%3D3.83%20%5Ctimes%2010%5E9%20%5C%20m%5C%5C%5C%5C)
a) ![48000 m^3](https://tex.z-dn.net/?f=48000%20m%5E3)
The volume of the sports hall can be calculated using the equation
![V=a\cdot b \cdot c](https://tex.z-dn.net/?f=V%3Da%5Ccdot%20b%20%5Ccdot%20c)
where
a, b, c are the measures of the sizes of the hall
For the sport hall in this problem, we have
a = 80 m
b = 40 m
c = 15 m
Substituting into the equation, we find
![V=(80)(40)(15)=48,000 m^3](https://tex.z-dn.net/?f=V%3D%2880%29%2840%29%2815%29%3D48%2C000%20m%5E3)
b) ![62400 kg](https://tex.z-dn.net/?f=62400%20kg)
Being a gas, the air fills the whole sport halls, therefore its volume is equal to the volume of the sports hall.
The relationship between the mass, the volume and the density of the air is
![\rho = \frac{m}{V}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7Bm%7D%7BV%7D)
where
is the density
m is the mass
V is the volume
Here we have:
![\rho = 1.3 kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%201.3%20kg%2Fm%5E3)
![V=48000 m^3](https://tex.z-dn.net/?f=V%3D48000%20m%5E3)
Solving for m, we find the mass of air:
![m=\rho V = (1.3)(48000)=62,400 kg](https://tex.z-dn.net/?f=m%3D%5Crho%20V%20%3D%20%281.3%29%2848000%29%3D62%2C400%20kg)