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skad [1K]
3 years ago
9

Outline the process used to test the following hypothesis: Titanium cages are stronger than steel cages for hockey goalie masks.

Engineering
1 answer:
son4ous [18]3 years ago
8 0

Answer:

true

Explanation:

You might be interested in
Pilot testing:
andrew11 [14]

Answer:

c. allows planners to work out any problems before the program is launched

Explanation:

Pilot testing is simply aimed at getting it right before the launch of a program, it is also called pilot run, pilot project, feasibility run, etc. Pilot testing is the rehearsal or practice done for an idea, program, research study or invention with few participants prior to lunching out the main program. The main purpose of pilot testing is to determine how feasible a project is, it can also help to evaluate the cost of an idea, invention, research study, etc.

6 0
3 years ago
A thin plastic membrane is used to separate helium from a gas stream. Under steady-state conditions the concentration of helium
Juli2301 [7.4K]

Answer:

N_A=1.5*10^-8 kmol/s.m^2

Explanation:

<u>KNOWN: </u>

Molar concentration of helium at the inner and outer surfaces of a plastic membrane. Diffusion coefficient and membrane thickness.  

<u>FIND:</u>

Molar diffusion flux.  

<u>ASSUMPTIONS:</u>

(1) Steady-state conditions, (2) One-dimensional diffusion in a plane wall, (3) Stationary medium, (4) Uniform C = C_A + C_B.  

<u>ANALYSIS:</u> The molar flux may be obtained from

 N_A=D_AB/L(C_A,1-C_A,2)

       =10^-9 m^2/s/0.001 m(0.02-0.005)kmol/m^3

N_A=1.5*10^-8 kmol/s.m^2

<u>COMMENTS:</u> The mass flux is:

n_A,x=M_a*N_A,x

n_A,x=6*10^-8 kg/s m^2

5 0
3 years ago
How many steps is a player allowed to take after catching a ball while running
KATRIN_1 [288]

Answer:

A moving player may only take 2 legal steps after catching the ball to shoot, pass or to come to a stop.

so (B. 2)

Explanation:

On this play, the offensive player catches the ball in the air and lands with a left, right, left prior to shooting. This is a traveling violation for taking 3 steps.

7 0
3 years ago
As the temperature of a thermal radiator is increased Group of answer choices the object appears redder. the object appears blue
Elanso [62]

Answer:

As the temperature of a thermal radiator is increased

Group of answer choices

  • the object appears redder.
  • the object appears bluer.
  • the object emits more power for the same area.
  • the object emits less power for the same area.
  • the object expands to keep the same power per units area.

<em>When the temperature of a thermal radiator increases ;</em>

  • <em>the object emits more power for the same are</em>
  • <em>the object becomes bluer</em>

Explanation:

Thermal radiation involves the transfer of heat between molecules of two substances without direct contact with each other. When a body is heated to a given temperature it begins to emit light which is transferred to nearby objects as thermal radiation. The medium through which the heat is transferred could be liquid, solid, or in a vacuum.

<h3>How temperature affects thermal radiation.</h3>

Temperature determines the amount of heat that is been radiated from a body. An increase in temperature would increase the thermal radiation of the body. The increase in the heat radiation results to increase in the thermal energy of the body. Also when a body is heated it tends to be bluer than a cool object, this is caused by the rapid movement of the molecules.

Therefore When the temperature of a thermal radiator increases ;

  • the object emits more power for the same are
  • the object becomes bluer
4 0
3 years ago
A small submarine has a triangular stabilizing fin on its stern. The fin is 1 ft tall and 2 ft long. The water temperature where
Arturiano [62]

Answer:

\mathbf{F_D \approx 1.071 \ lbf}

Explanation:

Given that:

The height of a  triangular stabilizing fin on its stern is 1 ft tall

and it length is 2 ft long.

Temperature = 60 °F

The objective is to determine the drag on the fin when the submarine is traveling at a speed of 2.5 ft/s.

From these information given; we can have a diagrammatic representation describing how the  triangular stabilizing fin looks like as we resolve them into horizontal and vertical component.

The diagram can be found in the attached file below.

If we recall ,we know that;

Kinematic viscosity v = 1.2075 \times 10^{-5} \ ft^2/s

the density of water ρ = 62.36 lb /ft³

Re_{max} = \dfrac{Ux}{v}

Re_{max} = \dfrac{2.5 \ ft/s \times 2  \  ft }{1.2075 \times 10 ^{-5} \ ft^2/s}

Re_{max} = 414078.6749

Re_{max} = 4.14 \times 10^5 which is less than < 5.0 × 10⁵

Now; For laminar flow;  the drag on  the fin when the submarine is traveling at 2.5 ft/s can be determined by using the expression:

dF_D = (\dfrac{0.664 \times \rho  \times U^2 (2-x) dy}{\sqrt{Re_x}})^2

where;

(2-x) dy = strip area

Re_x = \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}

Therefore;

dF_D = (\dfrac{0.664 \times 62.36  \times 2.5^2 (2-x) dy}{\sqrt{ \dfrac{2.5(2-x)}{1.2075 \times 10 ^{-5}}}})

dF_D = 1.136 \times(2-x)^{1/2} \ dy

Let note that y = 0.5x from what we have in the diagram,

so , x = y/0.5

By applying the rule of integration on both sides, we have:

\int\limits \  dF_D =  \int\limits^1_0 \  1.136 \times(2-\dfrac{y}{0.5})^{1/2} \ dy

\int\limits \  dF_D =  \int\limits^1_0 \  1.136 \times(2-2y)^{1/2} \ dy

Let U = (2-2y)

-2dy = du

dy = -du/2

F_D =  \int\limits^0_2 \  1.136 \times(U)^{1/2} \ \dfrac{du}{-2}

F_D = - \dfrac{1.136}{2} \int\limits^0_2 \ U^{1/2} \ du

F_D = -0.568 [ \dfrac{\frac{1}{2}U^{ \frac{1}{2}+1 }  }{\frac{1}{2}+1}]^0__2

F_D = -0.568 [ \dfrac{2}{3}U^{\frac{3}{2} }   ] ^0__2

F_D = -0.568 [0 -  \dfrac{2}{3}(2)^{\frac{3}{2} }   ]

F_D = -0.568 [- \dfrac{2}{3} (2.828427125)}   ]

F_D = 1.071031071 \ lbf

\mathbf{F_D \approx 1.071 \ lbf}

8 0
3 years ago
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