Convection is the movement<span> of groups of molecules within </span>fluids<span> such as gases and liquids, including molten rock (rheid).</span>
The velocity of the pitcher at the given mass is 0.1 m/s.
The given parameters:
- <em>Mass of the pitcher, m₁ = 50 kg</em>
- <em>Mass of the baseball, m₂ = 0.15 kg</em>
- <em>Velocity of the ball, u₂ = 35 m/s</em>
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Let the velocity of the pitcher = u₁
Apply the principle of conservation of linear momentum to determine the velocity of the pitcher as shown below;
m₁u₁ = m₂u₂

Thus, the velocity of the pitcher at the given mass is 0.1 m/s.
Learn more about conservation of linear momentum here: brainly.com/question/13589460
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:
The magintude of the acceleration for both objects is 
Explanation:
Drawing a free body diagram on the two boxes we can analyze the system more easily.
we can take the acceleration going up as positive for reference purposes.
for mA let's suppose that is ascending so:

and for mB (descending):


because the two boxes has the same acceleration because they are attached together:

So the magintude of the acceleration for both objects is 
Answer:
5
Explanation:
The d subshell has 5 orbitals, each capable of holding a maximum of two electrons. Hund's rule tells us that every orbital in a sub-level must first be singly occupied by electrons before any orbital is doubly occupied. Therefore five electrons will fill the five orbitals within the d subshell.