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Tom [10]
3 years ago
10

Solving 2-D Motion

Physics
2 answers:
Lesechka [4]3 years ago
3 0

do you have a formulai can do this problem

zavuch27 [327]3 years ago
3 0

I don't claim to be an expert here, but if I am correct the Delta Y is .434

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If a freely falling object were equipped with a speedometer on a planet where the acceleration due to gravity is 20 m/s², then i
babymother [125]

Answer:

20 m/s

Explanation:

The acceleration of an object determines the increase in velocity per second.

In this case the acceleration is 20m/s^2.

a=\frac{v}{t}=\frac{\frac{20m}{s}}{s}

That is, the increase is 20m/s each second.

I hope this is useful for you

Regards

7 0
4 years ago
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In scientific notation, (6.2 x 10^4) x (3.3 x 10^2 ) equals ____________________.
castortr0y [4]
You will get 20460000 as your answer which is broken down into, 2.046 x 10^7 as your number has to be between 1-10.
5 0
3 years ago
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Which result occurs during an exothermic reaction?
baherus [9]

Answer:

Explanation:

Chemical reactions that release energy are called exothermic. In exothermic reactions, more energy is released when the bonds are formed in the products than is used to break the bonds in the reactants. Exothermic reactions are accompanied by an increase in temperature of the reaction mixture.

3 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
A 13500 kg jet airplane is flying through some high winds. At some point in time, the airplane is pointing due north, while the
Mekhanik [1.2K]

Answer

given,

mass of the jet airplane = 13500 kg

Force on the plane = 35700 N due north

force from wind  = 15300 N in direction 80.0° south of west.

Force = 35700 \vec{j} N

force by wind = 15300(-cos \theta \vec{i}-sin \theta \vec{j}) N

                       = 15300(-cos 80^0 \vec{i}-sin 80^0 \vec{j}) N

net force on the jet airplane(ma)

          m a = 35700 \vec{j} + 15300(-cos 80^0 \vec{i}-sin 80^0 \vec{j})

          \vec{a} = \dfrac{35700}{13500} \vec{j} + \dfrac{15300}{13500}(-cos 80^0 \vec{i}-sin 80^0 \vec{j})

          \vec{a} = 2.64\vec{j} -0.197 \vec{i} - 1.116 sin 80^0 \vec{j})

           \vec{a} = -0.197 \vec{i} + 1.524 \vec{j}

a = \sqrt{-0.197^2+1.524^2}

a = 1.54 m/s²

\theta = tan^{-1}(\dfrac{-1.524}{0.197})

\theta = -82.63^0

3 0
3 years ago
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