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andre [41]
3 years ago
13

Future space stations will create an artificial gravity by rotating. Consider a cylindrical space station 780 m diameter rotatin

g about its central axis. Astronauts walk on the inside surface of the space station. What rotation period will provide "normal" gravity?
Physics
1 answer:
ser-zykov [4K]3 years ago
5 0

Answer:1.513 rps

Explanation:

Given

Diameter of cylindrical space d=780\ m

When the station rotates it creates centripetal acceleration which is given by

a_c=\omega ^2r

Now it must create the effect of gravity so

g=\omega ^2r

\omega =\sqrt{\frac{g}{r}}

\omega =0.158\ rad/s

and \omega =\frac{2\pi N}{60}

Thus N=\frac{4.755}{3.142}

N=1.513\ rps

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600,000

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do clouds and wind appear to follow the same patterns? Can you find any patterns in the direction that they move?
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6 0
4 years ago
48 You are given two amplifiers, A and B, to connect in cascade between a 10-mV, 100-k source and a 100- load. The amplifiers ha
Sedbober [7]

Answer:

SABL

Explanation:

The best amplifier will be the one that gives us a bigger gain. In each stage will be a load factor that will reduce the gain, that is defined as:

Fp=\frac{R_{in}}{R_{in}+R_{out}}\\

where Rin is the input resistance of the next stage and Rout the output resistance of the previous stage.

Analyzing SABL:

Fp_1=\frac{100K}{100K+100K}=0.5\\\\Fp_2=\frac{10K}{10K+10K}=0.5\\\\Fp_3=\frac{100}{100+1K}=0.0909

the total gain will be the total gain of each stage multiplied by the load factor.

SABL_{gain}=0.5*100*0.5*10*0.0909\\SABL_{gain}=22.73\\SABL_{gain_{db}}=20*\log{22.73}=27.13db

Analyzing SBAL:

Fp_1=\frac{10K}{10K+100K}=0.0909\\\\Fp_2=\frac{100K}{100K+10K}=0.909\\\\Fp_3=\frac{100}{100+10K}=0.0099

the total gain will be the total gain of each stage multiplied by the load factor.

SABL_{gain}=0.0909*10*0.99*100*0.0099\\SABL_{gain}=0.89\\SABL_{gain_{db}}=20*\log{0.89}=-1.01db

So the best amplifier arrangement is SABL.

4 0
3 years ago
A 37.5 kg box initially at rest is pushed 4.05 m along a rough, horizontal floor with a constant applied horizontal force of 150
vodomira [7]

Answer:

a) 607.5 J

b) 160.531875 J

c)  0 J

d)  0 J

e) 2.925 m\s

Explanation:

The given data :-

  • Mass of the box ( m ) = 37.5 kg.
  • Displacement made by box ( x ) = 4.05 m.
  • Horizontal force ( F ) = 150 N.
  • The co-efficient of friction between box and floor ( μ ) = 0.3
  • Gravitational force ( N ) = m × g = 37.5 × 9.81 = 367.875

Solution:-

a) The work done by applied force ( W )

W = force applied × displacement = 150 × 4.05 = 607.5 J

b)  The increase in internal energy in the box-floor system due to friction.

Frictional force ( f ) = μ × N = 0.3 × 367.875 = 110.3625 N

Change in internal energy = change in kinetic energy.

ΔU = ( K.E )₂ - ( K.E )₁

Since the initial velocity is zero so the  ( K.E )₁ = 0  

ΔU = ( K.E )₂ = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

c) The work done by the normal force .

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

d)  The work done by the gravitational force.

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

e) The change in kinetic energy of the box

( K.E )₂ - ( K.E )₁ = ( K.E )₂ - 0 = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

f) The final speed of the box

( K.E )₂ = 160.531875 J = 0.5 × 37.5 × v²

v² = 8.56

v = 2.925 m\s.

5 0
4 years ago
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liberstina [14]

Answer:

2, sorry I'm late but my answer is right, I just took the quiz

4 0
3 years ago
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