Answer:
(a). 14.4 lbf/in^2.
(b). 27.8 in, AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.
Explanation:
So, from the question above we are given the following parameters which are going to help us in solving this particular Question;
=> The "barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a vacuum at the top of the barometer)"
=> "On a day when the temperature is 70oF, the mercury column height is 28.35 inches (corrected for thermal expansion)."
With these knowledge, let us delve right into the solution;
(a). The barometric pressure = water vapor pressure + acceleration due to gravity (ft/s^2) × water density(slug/ft^3) × {ft/12 in}^3 × [ height of mercury column + specific gravity of mercury × height of water column].
The barometric pressure= 0.363 + {(62.146) ÷ (12^3) × 390.6425}. = 14.4 lbf/in^2.
(b). { (13.55 × length of mercury) + 6.5 } × (62.15÷ 12^3) = 14.4 - 0.603.
Length of mercury = 27.8 in.
AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.
Answer:
b.peeps(meaning "people "or "the friends your hang out with")
Answer:
Option D.) 5.0 x 10^-4m
Explanation:
Data obtained from the question include:
f (frequency) = 6 x 10^11 Hz
v (velocity) = 3.00 x 10^8 m/s
λ (wavelength) =?
Using the equation v = λf, the wavelength can be obtained as illustrated below:
v = λf
λ = v/f
λ = 3.00 x 10^8/6 x 10^11
λ = 5 x 10^-4m
Therefore, the wavelength of the beam of the electromagnetic radiation is 5 x 10^-4m
Answer:
Reflection is when light bounces of an object. If the surface is smooth and shiny, like glass, water or polished metal, the light will reflect at the same angle as it hit the surface.
Answer:
v = 60.7 m / s
Explanation:
This is an exercise that we can solve using the moment equations.
To begin we must create a system that is formed by the child and the block. For this system the forces during the crash are internal, for local moment it is conserved. Let's write the moment before and after the crash
Before
p₀ = m v₀ + 0
After
= (m + M) v
Where m and M are the masses of the child and the block, respectively. Notice that when they collide they are joined by which the mass is the sum of the two
p₀ = pf
m v₀ = (m + M) v
v = v₀ m / (m + M) (1)
We must find the speed of the child at 1.60 m, for this we use the law of conservation of energy, calculate the energy at two points: where it jumps from the trampoline and just before grabs the block
When he jumps off the trampoline
Em₁ = K = ½ m v₁²
Just before taking the block
Em₂ = K + U = ½ m v₂² + mg y
Em₁ = Em₂
½ m v₁² = ½ m v₂² + mg y
Let's calculate the speed with which it reaches the point of the crash
v₂² = v₁² - 2 g y
v₂ = √ (v₁² - 2 g y)
v₂ = √ (10.2² - 2 9.8 1.60)
v₂ = 72.7 m / s
This is the speed with which it reaches the point where the block is, vo = v₂ = 72.7 m/s, with this value we can substitute and calculate in equation (1) of conservation of the moment
v = vo m / (m + M)
v = 72.7 30.5 / (30.5 + 6.00)
v = 60.7 m / s
