Answer:
233.1 miles per hours
Explanation:
Speed: This is defined as the ratio of distance to time. The S.I unit of speed is m/s. speed is a vector quantity because it can only be represented by magnitude only. Mathematically, speed can be expressed as,
S = d/t ....................... Equation 1
Where S = speed of the runner, d = distance covered, t = time.
Given: d = 100 meter , t = 9.580 seconds
Conversion:
If, 1 meter = 0.00062 miles
Then, 100 meters = (0.00062×100) miles = 0.62 miles.
Also
If, 3600 s = 1 h
Then, 9.580 s = (1×9.580)/3600 = 0.00266 hours.
Substitute into equation 1
S = 0.62/0.00266
S = 233.1 miles per hours.
Hence the runner speed is 233.1 miles per hours
Answer:
²₁H + ³₂He —> ⁴₂He + ¹₁H
Explanation:
From the question given above,
²₁H + ³₂He —> __ + ¹₁H
Let ⁿₐX be the unknown.
Thus the equation becomes:
²₁H + ³₂He —> ⁿₐX + ¹₁H
We shall determine, n, a and X. This can be obtained as follow:
For n:
2 + 3 = n + 1
5 = n + 1
Collect like terms
n = 5 – 1
n = 4
For a:
1 + 2 = a + 1
3 = a + 1
Collect like terms
a = 3 – 1
a = 2
For X:
n = 4
a = 2
X =?
ⁿₐX => ⁴₂X => ⁴₂He
Thus, the balanced equation is
²₁H + ³₂He —> ⁴₂He + ¹₁H
Answer:
The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus.
Explanation:
The average speed of gas molecules is given by:
R is the gas constant, T is the temperature and M the molar mass of the gas.
We know that a water molecule has a mass that is 18 times that of a hydrogen atom:
So, we have:
The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus:
Answer:
a) P = 1240 lb/ft^2
b) P = 1040 lb/ft^2
c) P = 1270 lb/ft^2
Explanation:
Given:
- P_a = 2216.2 lb/ft^2
- β = 0.00357 R/ft
- g = 32.174 ft/s^2
- T_a = 518.7 R
- R = 1716 ft-lb / slug-R
- γ = 0.07647 lb/ft^3
- h = 14,110 ft
Find:
(a) Determine the pressure at this elevation using the standard atmosphere equation.
(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.
(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.
Solution:
- The standard atmospheric equation is expressed as:
P = P_a* ( 1 - βh/T_a)^(g / R*β)
(g / R*β) = 32.174 / 1716*0.0035 = 5.252
P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252
P = 1240 lb/ft^2
- The air density method which is expressed as:
P = P_a - γ*h
P = 2116.2 - 0.07647*14,110
P = 1040 lb/ft^2
- Using constant temperature ideal gas approximation:
P = P_a* e^ ( -g*h / R*T_a )
P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )
P = 1270 lb/ft^2
Light travels faster than sound ,because sound can only travel waves.
