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swat32
3 years ago
7

Look at the picture and help me plz

Physics
2 answers:
Sedaia [141]3 years ago
7 0

Answer:

AI

Explanation:

Elenna [48]3 years ago
5 0

Answer:

AI

Explanation:

it stands for artificial intelligents

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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
Our molecular model of matter describes a fluid as consisting of lots of little particles (atoms or molecules)
sergij07 [2.7K]

Answer:

a)  μ = [kg / m s]

b)  D = m²/s

Explanation:

Viscosity is equivalent to a friction force that opposes the movement of fluids, it is defined by the relation

       

       

      Shear stress = μ speed gradient

       Fr/A = μ  v₀ / h

Where Fr is the shear force, A the area, h the height and v₀ the velocity of the fluid layer

     μ = [kg / m s]

The diffusion coefficient characterizes the ability of a material to move in a given solvent, depending on the size of the solute, the viscosity of the solvent, temperature

           

         J = - D dΘ / dx

J is the broadcast glare

            D = [m² / s]

5 0
3 years ago
A 750 kg National Geographic Drone is rising vertically up into the atmosphere at constant speed. The lift force pushing it upwa
Debora [2.8K]

Answer:

a. 0

b. 1.103625 MJ

c. Conserved

d. 1.103625/n MJ where n = The number of forces

Explanation:

The mass of the drone, m = 750 kg

The upward lift force = 125% of the weight of the drone

The time it takes the drone to reach a height of 250 m = 25 seconds

a. The mechanical energy = The kinetic energy + Potential energy

Therefore, given that the drone stars motion from the surface and was initially at rest, the mechanical energy at the surface = 0

b. The mechanical energy at height, h = 150 m, ME₁₅₀ = The potential energy gained = m·g·h

Where;

g = The acceleration due to gravity = 9.81 m/s²

∴ ME₁₅₀ = 750 kg × 9.81 m/s² × 150 m = 1103625 J = 1.103625 MJ

c. The mechanical energy is equivalent to the potential energy of the drone at the 150 m height, therefore, it is conserved

d. The work done by the force = The energy gained

Therefore, where there are <em>n</em> number of forces, the work done by each force = 1.103625/n MJ

5 0
3 years ago
An electron moves in a region where the magnetic field is uniform and has a magnitude of 80 μT. The electron follows a helical p
sladkih [1.3K]

Answer:

3.4 x 10⁴ m/s

Explanation:

Consider the circular motion of the electron

B = magnetic field = 80 x 10⁻⁶ T

m = mass of electron = 9.1 x 10⁻³¹ kg

v  = radial speed

r = radius of circular path = 2 mm = 0.002 m

q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

For the circular motion of electron

qBr = mv

(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v

v = 2.8 x 10⁴ m/s

Consider the linear motion of the electron :

v' = linear speed

x = horizontal distance traveled = 9 mm = 0.009 m

t = time taken = \frac{2\pi m}{qB} = \frac{2\pi (9.1\times 10^{-31})}{(1.6\times 10^{^{-19}})(80\times 10^{-6})} = 4.5 x 10⁻⁷ sec

using the equation

x = v' t

0.009 = v' (4.5 x 10⁻⁷)

v' = 20000 m/s

v' = 2 x 10⁴ m/s

Speed is given as

V = sqrt(v² + v'²)

V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)

v = 3.4 x 10⁴ m/s

6 0
4 years ago
Calculate the frequency if the number of revolutions is 300 and the paired poles are 50.
Andrei [34K]

Answer: A

Explanation: We know that f=p*n

f=50*300=15000 Hz = 15kHz.

Have a great day! <3

3 0
2 years ago
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