The correct answer should be c.The kinetic energy of the water molecules decreases.
If the temperature drops that means that the molecules are coming together. If the temperature rises then it means that the molecules are spreading. If the kinetic energy falls down that means that they are slower which means that they are cooler.
Answer: The riders are subjected to 11.5 revolutions per minute
Explanation: Please see the attachments below
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
Explanation:
<u>Mass of car</u> = 137.5 kg
<u>Acceleration</u> = v - u / t = 26 - 0 / 6 = 4.33 m/sec^2
Force = m * a = 137.5 * 4.33 = 595.3 N
Answer:
2.47 m
Explanation:
Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.
The horizontal velocity of the ball is constant:
and the time taken to cover the horizontal distance d is
So this is the time the ball takes to reach the horizontal position of the crossbar.
The vertical position of the ball at time t is given by
where
is the initial vertical velocity
g = 9.8 m/s^2 is the acceleration of gravity
And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:
The height of the crossbar is h = 3.05 m, so the ball passes
above the crossbar.
