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frosja888 [35]
3 years ago
10

The electric field that is 0.25m from a small sphere is 450n/c toward the sphere.

Physics
2 answers:
Andrei [34K]3 years ago
8 0
Its b bc its the volume of the sphere 
Hitman42 [59]3 years ago
5 0

PART 1)

A small sphere is considered to be a point charge

So here if electric field is indicating the position of charge itself or it is towards the charge then it must be a negative charge on the sphere

PART 2)

Electric field due to a point charge is given by

E = \frac{KQ}{r^2}

here

k = 9 * 10^9

Q = charge

r = distance of charge from the point where field strength is required

= 0.25 m

E = 450 N/c

now from the above equation we can say

450 = \frac{9*10^9 * Q}{0.25^2}

Q = 3.125 * 10^{-9} C

So magnitude of charge is 3.125 nC on the small sphere

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a race car accelerates uniformly from 18.5 m/s to 46.1m/s in 2.47 seconds detrrmine the acceleration of the car and distance tra
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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
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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

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

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\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

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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
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