You could solve this in two ways: using the ideal gas law (Van der Waals parameters optional), or by using the density. Since you specify a pressure and temperature (in kelvin), I will use the ideal gas law. Ideal gas law: PV=nRT P = pressure = 1.00 ATM V = volume = 1.00 L n = moles = what you're solving for R = gas constant = 0.0821 L*ATM/(mol*T) (T is absolute temperature (kelvin)) T = absolute temperature = 298 K (1.00atm)(1.00L) = n(0.0821L*ATM*mol -1 T -1 )(298K) n = 0.04 moles A n = Avogadro's Number = number of molecules in one mole = 6.022141 * 10 23 0.04 * 6.022141*10 23 = 2.409 * 10 22 molecules of N 2 in 1.00L at 1.00atm and 298K
Answer
As the angle of incidence increases in Figure 2.8, a point is finally reached where the refracted ray does not emerge at the second layer but lie along the interface. This particular angle of incidence at which the angle of refraction is 90° and the refracted ray lies along the interface is known as the critical angle. At and beyond the critical angle, there is no transmitted ray and therefore a very high reflected ray will be recorded .
Therefore,
sinθisin90=Vp1Vp2
But, sin 90 = 1.
At critical angle,
sinθcritical=Vp1Vp2
A critical refracted wave travels along the interface between layers and is refracted back into the upper layer at the critical angle. The waves refracted back into the upper layer are called head waves or first-break refractions because at certain distances from a source, they are the first arriving energy. Recorded first-break refraction is shown in Figure 2.10.
Note that these first-break refractions can give us important information about the shallow velocities on land seismic data.
Note also that seismic data are acquired in such a way that reflections from horizons of interest are in the pre-critical region, even at the farthest offset in the data.
In reality, part of the seismic energy arriving at an interface is transmitted and refracted, and another part of the energy is reflected at that same interface. Given that there are many reflectors in the subsurface, there will be many paths from source to receiver, each of them with a different travel time. The proportion of energy reflected depends on the material properties of the two bounding layers and on the angle of incidence
Answer:
f(-9) = 184
Explanation:
f(x)=3x²+5x-14
f(-9)= 3(-9)² +5(-9)-14 Order of Operations : Exponents
= 3(81)+5(-9)-14
= 243+5(-9)-14
= 243-45-14
= 198-14
f(-9)= 184
Hope this helps :)
Answer:
v = 12.52 [m/s]
Explanation:
To solve this problem we must use the energy conservation theorem. Which tells us that potential energy is transformed into kinetic energy or vice versa. This is more clearly as the potential energy decreases the kinetic energy increases.
Ep = Ek
where:
Ep = potential energy [J] (units of joules]
Ek = kinetic energy [J]
Ep = m*g*h
where:
m = mass of the rock = 45 [g] = 0.045 [kg]
g = gravity acceleration = 9.81 [m/s²]
h = elevation = (20 - 12) = 8 [m]
Ek = 0.5*m*v²
where:
v = velocity [m/s]
The reference level of potential energy is taken as the ground level, at this level the potential energy is zero, i.e. all potential energy has been transformed into kinetic energy. In such a way that when the Rock has fallen 12 [m] it is located 8 [m] from the ground level.
m*g*h = 0.5*m*v²
v² = (g*h)/0.5
v = √(9.81*8)/0.5
v = 12.52 [m/s]
Answer:
The equation of motion is derived based on the Newton’s laws of motion. And it changes accordingly when an object changes with uniform velocity.
Given is that object moves with uniform velocity, that is no change in velocity so there will no acceleration.
As we know
Here, u = v (due to uniform velocity)
.
1st equation of motion is, v = u + at
2nd equation of motion,
3rd equation of motion,
.
Explanation:
