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2 years ago

How many steps is a player allowed to take after catching a ball while running

Engineering

1 answer:

KATRIN_1 [288]2 years ago

7 0

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.

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A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe

ankoles [38]

Answer:

F = 8849 N

Explanation:

Given:

Load at a given point = F = 4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ $_{fs} = F_{f} L / \pi R^{3}$

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π ( 5.6 x 10⁻³ ) ³

= 4250 ( 44 x 10⁻³ ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 ( 11 / 250 ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

= 187 / 3.141593 (0.0056)³

= 338943767.745358

= 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

$F_{f}$ = 2 σ $_{fs}$ d³/3 L

F = 2σd³/3L

= 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

= 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 1/1000 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 3 / 250 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 27 / 15625000 ) / 3 ( 44 x 10⁻³)

= 146016 / 125 / 3 ( 44 x 1 / 1000 )

= ( 146016 / 125 ) / (3 ( 11 / 250 ))

= 97344 / 11

F = 8849 N

4 0

2 years ago

What type of engineer would be most likely to develop a design for cars? chemical civil materials mechanical

Andreyy89

I don’t know but good luck

4 0

2 years ago

Can crushers help us recycle in a space efficient way which is good for saving the earth and for giving you more room in your ap

777dan777 [17]

Answer:

Force magnitude = 296.7 N

Explanation:

Detailed illustration is given in the attached document.

6 0

2 years ago

What oil specification is used to express the thickness (viscosity) of the oil?

Marianna [84]

Answer:

Viscosity is notated using the common classification “XW-XX”. The number preceding the “W” (winter) rates the oil's flow (viscosity) at zero degrees Fahrenheit (-17.8 degrees Celsius). The lower the number, the less the oil thickens in cold weather.

7 0

1 year ago

Water vapor at 10bar, 360°C enters a turbine operatingat steady state with a volumetric flow rate of 0.8m3/s and expandsadiabati

Artyom0805 [142]

Answer:

A) W' = 178.568 KW

B) ΔS = 2.6367 KW/k

C) η = 0.3

Explanation:

We are given;

Temperature at state 1;T1 = 360 °C

Temperature at state 2;T2 = 160 °C

Pressure at state 1;P1 = 10 bar

Pressure at State 2;P2 = 1 bar

Volumetric flow rate;V' = 0.8 m³/s

A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;

Specific volume;v1 = 0.287322 m³/kg

Mass flow rate of water vapour at turbine is defined by the formula;

m' = V'/v1

So; m' = 0.8/0.287322

m' = 2.784 kg/s

Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific enthalpy;h1 = 3179.46 KJ/kg

Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific enthalpy;h2 = 3115.32 KJ/kg

Now, since stray heat transfer is neglected at turbine, we have;

-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2 - h1)

Plugging in relevant values, the work of the turbine is;

W' = -2.784(3115.32 - 3179.46)

W' = 178.568 KW

B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific entropy: s1 = 7.3357 KJ/Kg.k

Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific entropy; s2 = 8.2828 KJ/kg.k

The amount of entropy produced is defined by;

ΔS = m'(s2 - s1)

ΔS = 2.784(8.2828 - 7.3357)

ΔS = 2.6367 KW/k

C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;

h2s = 2966.14 KJ/Kg

Energy equation for turbine at ideal process is defined as;

Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Again, Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2s - h1)

W' = -2.784(2966.14 - 3179.46)

W' = 593.88 KW

the isentropic turbine efficiency is defined as;

η = W_actual/W_ideal

η = 178.568/593.88 = 0.3

8 0

2 years ago