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k0ka [10]
3 years ago
10

A 50 kg astronaut floating in space throws her 2 kg wrench to the left at 10

Physics
1 answer:
Hoochie [10]3 years ago
8 0

The astronaut will move at 0.4 m/s in the opposite direction to the wrench

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the astronaut-wrench system must be conserved before and after the launch.

Before the launch, the total momentum is zero, since the astronaut is at rest:

p = 0 (1)

After the launch, the total momentum is:

p=mv+MV (2)

where :

m = 2 kg is the mass of the wrench

v = 10 m/s is the velocity of the wrench

M = 50 kg is the mass of the astronaut

V is the recoil velocity of the astronaut

Since momentum is conserved, we can write (1) = (2), and so we can solve for V:

0=mv+MV\\V=-\frac{mv}{M}=-\frac{(2)(10)}{50}=-0.4 m/s

And the negative sign means that the astronaut will move in the opposite direction to the wrench.

Learn more about conservation of momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

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Answer:

a) The shear stress is 0.012

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Explanation:

Given by the problem:

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v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

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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

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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

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