As a basketball ball player starts to jump for rebound

Does the atmosphere contain 21% nitrogen in 78% oxygen

Alex777 [14]

Answer:

An atmosphere is the layers of gases surrounding a planet or other celestial body. Earth's atmosphere is composed of about 78% nitrogen, 21% oxygen, and one percent other gases

7 0

2 years ago

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

7 0

3 years ago

What is Tension variables?

Soloha48 [4]

Answer:

The tension on an object is equal to the mass of the object x gravitational force plus/minus the mass x acceleration. T = mg + ma.

Explanation:

6 0

2 years ago

You collect some more data on that horse at a later time interval, but now you are measuring thehorse’s velocity, not its positi

Monica [59]

Answer:

a) x(t) = 10t + (2/3)*t^3

b) x*(0.1875) = 10.18 m

Explanation:

Note: The position of the horse is x = 2m. There is a typing error in the question. Otherwise, The solution to cubic equation holds a negative value of time t.

Given:

- v(t) = 10 + 2*t^2 (radar gun)

- x*(t) = 10 + 5t^2 + 3t^3 (our coordinate)

Find:

-The position x of horse as a function of time t in radar system.

-The position of the horse at x = 2m in our coordinate system

Solution:

- The position of horse according to radar gun:

v(t) = dx / dt = 10 + 2*t^2

- Separate variables:

dx = (10 + 2*t^2).dt

- Integrate over interval x = 0 @ t= 0

x(t) = 10t + (2/3)*t^3

- time @ x = 2 :

2 = 10t + (2/3)*t^3

0 = 10t + (2/3)*t^3 + 2

- solve for t:

t = 0.1875 s

- Evaluate x* at t = 0.1875 s

x*(0.1875) = 10 + 5(0.1875)^2 + 3(0.1875)^3

x*(0.1875) = 10.18 m

3 0

3 years ago

If you heat soup on a stove, what happens to the movement of the soup’s particles?

deff fn [24]

Answer: B. The particles mover faster.

6 0

3 years ago

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