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avanturin [10]
3 years ago
14

Would u rather/ be able to fly. or be able to turn invisable

Physics
2 answers:
Tcecarenko [31]3 years ago
8 0
I would rather turn invisible because you can sneak into places witch are restricted and see what kind of things they are doing personally for me because I want to see what and how they built and design space aircraft and lien from the best face to face since In the future I want to become an Aerospace Engineer.
mojhsa [17]3 years ago
4 0

Answer:

fly......................

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An airplane travels 2100 km at a speed of 1000 km/h. It then encounters a headwind which slows it to 800 km/h for the next 1300
Sidana [21]
I don’t get it I need help
5 0
3 years ago
A bird sits on a high-voltage power line with its feet 3.87 cm apart. The wire is made from aluminum, is 2.11 cm in diameter, an
Svetlanka [38]

Answer:

ΔV=0.484mV

Explanation:

The potential difference across the end of conductor that obeys Ohms law:

ΔV=IR

Where I is current

R is resistance

The resistance of a cylindrical conductor is related to its resistivity p,Length L and cross section area A

R=(pL)/A

Given data

Length L=3.87 cm =0.0387m

Diameter d=2.11 cm =0.0211 m

Current I=165 A

Resistivity of aluminum p=2.65×10⁻⁸ ohms

So

ΔV=IR

=I(\frac{pL}{A})\\ =I(\frac{pL}{\pi r^{2} } )\\=I(\frac{pL}{\pi (d/2)^{2} } )\\=165A((\frac{(2.65*10^{-8})(0.0387m)}{\pi (0.0211m/2)^{2} } ))\\=4.84*10^{-4}V

ΔV=0.484mV  

3 0
3 years ago
A space station shaped like a giant wheel has a radius of a radius of 153 m and a moment of inertia of 4.16 × 10⁸ kg·m² (when it
Naya [18.7K]

Answer:

a = 1.709g

Explanation:

Given the absence of external forces being applied in the space station, it is possibly to use the Principle of Angular Momentum Conservation, which states that:

I_{o} \cdot \omega_{o} = I_{f} \cdot \omega_{f}

The required initial angular speed is obtained herein:

g= \omega_{o}^{2}\cdot R_{ss}

\omega_{o}=\sqrt{\frac{g}{R_{ss}} }

\omega_{o}= \sqrt{\frac{9.807\,\frac{m}{s^{2}} }{153\,m} }

\omega_{o} \approx 0.253\,\frac{rad}{s}

The initial moment of inertia is:

I_{o} =I_{ss}+n\cdot m_{person}\cdot R_{ss}^{2}

I_{o} = 4.16\times 10^{8}\,kg\cdot m^{2}+(150)\cdot (65\,kg)\cdot (153\,m)^{2}

I_{o} = 6.442\times 10^{8}\,kg\cdot m^{2}

The final moment of inertia is:

I_{f} =I_{ss}+n\cdot m_{person}\cdot R_{ss}^{2}

I_{f} = 4.16\times 10^{8}\,kg\cdot m^{2}+(50)\cdot (65\,kg)\cdot (153\,m)^{2}

I_{f} = 4.921\times 10^{8}\,kg\cdot m^{2}

Now, the final angular speed is obtained:

\omega_{f} = \frac{I_{o}}{I_{f}}\cdot \omega_{o}

\omega_{f} = \frac{6.442\times 10^{8}\,{kg\cdot m^{2}}}{4.921\times 10^{8}\,kg\cdot m^{2}} \cdot (0.253\,\frac{rad}{s} )

\omega_{f} = 0.331\,\frac{rad}{s^}

The apparent acceleration is:

a_{f} = \omega_{f}^{2}\cdot R_{ss}

a_{f} = (0.331\,\frac{rad}{s} )^{2}\cdot (153\,m)

a_{f} = 16.763\,\frac{m}{s^{2}}

This is approximately 1.709g.

4 0
4 years ago
2. If, in the previous question, there is a 15 N kinetic frictional force opposing the motion, how much work is
NeTakaya

Answer:

this answer your question

4 0
3 years ago
On a sky coaster (human pendulum) that reaches 10 meters from it's equilibrium position, a man of 120 kg is able to reach a maxi
dimaraw [331]

Answer:

14 m/s

Explanation:

Using the principle of conservation of energy, the potential energy is converted to kinetic energy, assuming any losses.

Kinetic energy is given by ½mv²

Potential energy is given by mgh

Where m is the mass, v is the velocity, g is acceleration due to gravity and h is the height.

Equating kinetic energy to be equal to potential energy then

½mv²=mgh

V

Making v the subject of the formula

v=√(2gh)

Substituting 9.81 m/s² for g and 10 m for h then

v=√(2*9.81*10)=14.0071410359145 m/s

Rounding off, v is approximately 14 m/s

6 0
4 years ago
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