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RSB [31]
3 years ago
10

An iron ball is dropped at a height of 10 m from the surface of the moon.

Physics
1 answer:
galben [10]3 years ago
3 0

Answer:

3.51s

Explanation:

There are many students who can not get answers step by step and on time

So there are a wats up group where you can get help step by step and well explained by the trusted experts.

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Match each situation
EastWind [94]
A: is potential
C: is losing kinetic energy and gaining potential energy
B: kinetic energy is at its highest
D: is loosing potential energy and gaining kinetic energy
3 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
2 years ago
Ohm’s law describes the relationship between which quantities?
VladimirAG [237]
I think its between b or c
5 0
3 years ago
Read 2 more answers
Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant
Elena L [17]

Answer:

19 m/s

Explanation:

The complete question requires the final speed to be calculated.

Velocity is the rate and direction at which an object moves. Acceleration is the rate of change of velocity per unit time and can be calculated by the difference in velocity over a given time.

For this question, first the unknown acceleration must be calculated and used to determine the final velocity

Step 1: Calculate the acceleration

a=\frac{v_{2}-v{1}}{t_{1}}

a=\frac{11-7}{8}

a=\frac{4}{8}

a=0.5 m/s^{2}

Step 2: Calculate the velocity using the acceleration calculated above

a=\frac{v_{3}-v{2}}{t_{2}}

0.5=\frac{v_{3}-11}{16}

v_{3}=19 m/s

3 0
3 years ago
Determine tge energy in joules of a photon whose frequency is 3.55x10
JulijaS [17]
E = hf
E = 6.63×10^-34 × 3.55×10 eV
1 eV = 1.60×10^-19 J
E = 6.63×10^-34 × 3.55×10 × 1.60×10^-19
E = 3.77×10^-51 J

Hope it helped!
6 0
3 years ago
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