Answer:
An investigation is made to determine the performance of simple thin airfoils in the slightly supersonic flow region with the aid of the nonlinear transonic theory first developed by von Kármán[1]. Expressions for the pressure coefficient across an oblique shock and a Prandtl-Meyer expansion are developed in terms of a transonic similarity parameter. Aerodynamic coefficients are calculated in similarity form for the flat plate and asymmetric wedge airfoils, and curves are plotted. Sample curves for a flat plate and a specific asymmetric wedge are plotted on the usual coordinate grid of Cl, Cd,andCmc/4versus angle of attack and Cl versus Mach Number to illustrate the apparent features of nonlinear flow.
Explanation:
Answer:
RIGHT
Explanation:
given,
mass of box = 250 kg
speed of the truck = 25 m/s
static friction µs = 0.7
g = 10 m/s²
to find direction of friction on the box
truck is moving in left direction and suddenly in front of truck a car stops then the driver hit the brake hard so at that moment the box on the trailer will have the tendency to move left so to oppose its movement friction force will be on RIGHT
The current I is defined as the amount of charge Q flowing through a certain point in a time t:
![I= \frac{Q}{t}](https://tex.z-dn.net/?f=I%3D%20%5Cfrac%7BQ%7D%7Bt%7D%20)
which can be rewritten as
![Q=It](https://tex.z-dn.net/?f=Q%3DIt)
so, by using this formula we can calculate the maximum amount of charge that can flow in the circuit. In fact, the maximum current is I=15 A, while the time is
![t=24 h=86400 s](https://tex.z-dn.net/?f=t%3D24%20h%3D86400%20s)
so the maximum amount of charge is
N₀ = number of radioactive atoms initially present = 128
N = number of radioactive atoms left after time "t" = 2
t = time taken
T = half life period for the radioactive metal = 20 years
λ = decay constant = ?
decay constant is given as
λ = 0.693/T
inserting the values
λ = 0.693/20
λ = 0.0347
number of radioactive atoms left after time "t" is given as
N = N₀ ![e^{-\lambda t}](https://tex.z-dn.net/?f=e%5E%7B-%5Clambda%20t%7D)
inserting the values
2 = (128) ![e^{-(0.0347) t}](https://tex.z-dn.net/?f=e%5E%7B-%280.0347%29%20t%7D)
t = 120 years
Answer:
a. The Sun is 110.24 cm.
b. The distance of the nearest star at 4.24 light years is 3.1586 ×
cm.
Explanation:
Given a scale :
12700 Km : 1 cm
a. How large is the Sun of diameter 1.4 ×
Km?
Thus using the given scale, we have to compare the diameter of the Sun to that of the Earth.
= ![\frac{1.4*10^{6} }{12700}](https://tex.z-dn.net/?f=%5Cfrac%7B1.4%2A10%5E%7B6%7D%20%7D%7B12700%7D)
= 110.2362
The Sun is 110.24 cm.
b. how far away is the nearest star at a distance of 4.24 light-years on this scale?
A light year is a term that is used to express the total distance that light would travel in an empty space in a year.
One light year = 9.461 ×
Km
So that,
4.24 light years = 4.24 × 9.461 ×
Km
= 40.11464 ×
Km
Using the given scale,
![\frac{40.11464*10^{12} }{12700}](https://tex.z-dn.net/?f=%5Cfrac%7B40.11464%2A10%5E%7B12%7D%20%7D%7B12700%7D)
= 3.1586 × ![10^{9}](https://tex.z-dn.net/?f=10%5E%7B9%7D)
The distance of the nearest star at 4.24 light years is 3.1586 ×
cm.