Approximately how old is the earth?

TRUE OR FALSE: The following drops were most likely dropped from a 90 degree angle.

horsena [70]

Answer:

True

Explanation:

If it weren't from a 90 degree angle then the circle would be a bit more oval shaped

7 0

3 years ago

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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

Diffusion occurs faster in gases than in liquids because _____.

jolli1 [7]

The correct answer to go in the blank would be A) The particles are moving faster.

5 0

3 years ago

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Ted William drops a ball from 14.5 meter to a desk that is 1.9 meters tall. What is the final speed of the ball right before it

makkiz [27]

Answer:

15.7m/s

Explanation:

To solve this problem, we use the right motion equation.

Here, we have been given the height through which the ball drops;

Height of drop = 14.5m - 1.9m = 12.6m

The right motion equation is;

V² = U² + 2gh

V is the final velocity

U is the initial velocity = 0

g is the acceleration due to gravity = 9.8m/s²

h is the height

Now insert the parameters and solve;

V² = 0² + 2 x 9.8 x 12.6

V² = 246.96

V = √246.96 = 15.7m/s

4 0

3 years ago

Check the correctness of formula t=2π √m/k, dimensionally

pantera1 [17]

Hi there! Lets see!

m is mass, and its units are kg
k is the elastic constant measured in newtons per meter (N/m), or kilograms per second squared kg/s²

Therefore:

$\sqrt{\dfrac{m}{k}} =\sqrt{\dfrac{[kg]}{[\dfrac{kg}{s^2}]}} =\sqrt{\dfrac{[kg]}{[kg]}\cdot s^2} = \sqrt{[s]^2} = s$

The period is given in seconds so the formula is dimensionally correct.

4 0

3 years ago