Answer:c
Explanation:
Given
object is falling Freely with an odometer
Suppose it falls with zero initial velocity
so distance fallen in time t is given by

here u=0 and t=time taken

for 

for 

distance traveled in 2 nd sec
for 

distance traveled in 3 rd sec
so we can see that distance traveled in each successive second is increasing
1.Paper Chromatography. This method is often used in the food industry. ...
2.Filtration. This is a more common method of separating an insoluble solid from a liquid. ...
3.Evaporation. ...
4Simple distillation. ...
Fractional distillation.
Answer:

Explanation:
Given

Required
Rewrite using scientific notation
The format of a number in scientific notation is

Where 
So the given parameter can be rewritten as

Express as a power of 10

Hence, the equivalent of the mass of the sun in scientific notation is:

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
The gravitational acceleration of a planet is proportional to the planet's mass, and inversely proportional to square of the planet's radius.
So when you stand on the surface of this particular planet, you feel a force of gravity that is
(1/2) / (3²)
of the force that you feel on the surface of the Earth.
That's <em>(1/18)</em> as much as on Earth.
The acceleration of gravity there would be about <em>0.545 m/s²</em>.
This is about 12% less than the gravity on Pluto.