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Monica [59]
3 years ago
11

An object has a velocity of 8 m/s and a kinetic energy of 480 j what is the mass of the object

Physics
2 answers:
dedylja [7]3 years ago
7 0

We Know,

K.E. = 1/2 mv²

480 = 1/2 (m)(8)²

m = 960/64

m = 15 Kg

So, the mass of the object is 15 Kg


Airida [17]3 years ago
7 0

Hello!

An object has a velocity of 8 m/s and a kinetic energy of 480 J. What is the mass of the object?

We have the following data:  

KE (Kinetic Energy) = 480 J

m (mass) = ? (in Kg)

v (speed) = 8 m/s  

Formula to calculate kinetic energy:  

KE = \dfrac{1}{2} *m*v^2

Solving:  

KE = \dfrac{1}{2} *m*v^2

480 = \dfrac{1}{2} *m*8^2

480 = \dfrac{1}{2} *m*64

480*2 = 64*m

960 = 64m  

64m = 960

m = \dfrac{960}{64}

\boxed{\boxed{m = 15\:Kg}}\end{array}}\qquad\checkmark

Answer:

15 kg

_______________________

I Hope this helps, greetings ... Dexteright02! =)

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Describe the difference between distance, position, and displacement
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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
3 years ago
Freeeeeeeeeeeeeeeeeeeeeeeee points
andreyandreev [35.5K]

Answer:

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3 0
2 years ago
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