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kakasveta [241]
3 years ago
9

Can a ratio be expressed as a percentage

Physics
1 answer:
cupoosta [38]3 years ago
7 0
Yes, ratio can be expressed in percentage.
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Idk I would go back to pre-k
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3 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
Mass, volume and density are all properties of
snow_tiger [21]

Answer:

Properties of matter

Explanation:

All properties of matter are either extensive or intensive and either physical or chemical. Extensive properties, such as mass and volume, depend on the amount of matter that is being measured. Intensive properties, such as density and color, do not depend on the amount of matter.

7 0
3 years ago
Read 2 more answers
In a transpiration experiment, the air bubble has an initial volume of 4.33 mL, and an initial pressure of 101.9 kPa. What will
ira [324]

Answer:

Explanation:is in the picture below

8 0
3 years ago
A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed
Vladimir [108]

Answer:

we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level

for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.

Explanation:

A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed of the rock at 2m on the way down compare with its speed at 2m on the way up?

It decreases in speed on its way down and increases in speed on its way down.

it decreases in speed on its way up because the the vertical motion is against the earths gravitational pull on an object to the earth's center

.It increases in speed on his way down because its under the influence of gravity

from newton's equation of motion we can check by

using V^2=u^2+2as

we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level

for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.

5 0
3 years ago
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