A raging activity can be found in t<span>he Sun's interior, with pressure waves being produced and travelling back and forth, from the core to the surface and back to the core. By looking closely at the 'surface' we can see these "ripples". It gives us an idea of how dense the material was that the waves passed through. In a way, this can help to predict solar storms in the future.</span>
Answer:
21 m
Explanation:
The motion of the frog is a uniform motion (constant speed), therefore we can find the distance travelled by using

where
d is the distance covered
v is the speed
t is the time
The frog in this problem has a speed of
v = 2.1 m/s
and therefore, after t = 10 s, the distance it covered is

Answer:
The polarity of water molecules means that molecules of water will stick to each other like when unlike charges attracts. This is called hydrogen bonding.
Polarity makes water a good solvent, gives it the ability to stick to itself (cohesion), stick to other substances (adhesion), and have surface tension (due to hydrogen bonding).
When the two hydrogen atoms bond with the oxygen, they attach to the top of the molecule. This molecular structure gives the water molecule polarity, or a lopsided electrical charge that attracts other atoms. The end of the molecule with the two hydrogen atoms is positively charged.
Explanation:
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
Answer:
Speed at which the ball passes the window’s top = 10.89 m/s
Explanation:
Height of window = 3.3 m
Time took to cover window = 0.27 s
Initial velocity, u = 0m/s
We have equation of motion s = ut + 0.5at²
For the top of window (position A)

For the bottom of window (position B)


We also have

Solving

So after 1.11 seconds ball reaches at top of window,
We have equation of motion v = u + at

Speed at which the ball passes the window’s top = 10.89 m/s