3 years ago

Can someone help me with this please?

Physics

1 answer:

shutvik [7]3 years ago

7 0

Answer:

Your answer will be C. 3.8 kPa.

You might be interested in

What is the displacement of this time velocity graph i need help asap

maw [93]

Answer:

b is the answer

Explanation:

because i did this before

3 0

3 years ago

The first step when doing<br> an<br> investigation is the observe a situation. True or false?

Sunny_sXe [5.5K]

Answer:

True! First step is to make objective observations.

6 0

3 years ago

Read 2 more answers

Determine if the data are qualitative or quantitative.

Amanda [17]

Answer:

Qualitative, Quantitative, Qualitative, Quantitative, and Qualitative.

Explanation:

7 0

2 years ago

Read 2 more answers

A force of 45 newtons is applied on an object, moving it 12 meters away in the same direction as the force. What is the magnitud

klio [65]

If the applied force is in the same direction as the object's displacement, the work done on the object is:

W = Fd

W = work, F = force, d = displacement

Given values:

F = 45N

d = 12m

Plug in and solve for W:

W = 45(12)

W = 540J

5 0

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