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2 years ago

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A space station consists of three modules, connected to form an equilateral triangle of side length 82.0 m. Suppose 100 people,

with an average mass of 75.0 kg each, live in each capsule and the mass of the modules is negligible compared to the mass of the people. At the current rotational rate the effective acceleration of gravity is g/2. (a) What angular momentum of the system

1 answer:

CaHeK987 [17]2 years ago

8 0

Answer:

Angular momentum of the system is 16221465.4617 kgm²/s

Explanation:

Given that;

length of the side of the triangle L = 82 M

m = 75.0 kg × 100 = 7500 kg

distance of each vertex from center R = L/√3 = 82/√3 = 47.34 m

effective acceleration a = 9.8 / 2 = 4.9 m/s²

we know that; effective acceleration is being provided by centripetal acceleration.

so

a = R × w²

rate of rotation w = √( a / R) = √( 4.9 / 47.34) = 0.3217 rad/seconds

Moment of Inertia I = 3mR²

we substitute

I = 3 × 7500 × (47.34)²

Also, Angular momentum L is expressed as;

L = I × w

so

L = 3 × 7500 × (47.34)² × 0.3217

L = 16221465.4617 kgm²/s

Therefore, Angular momentum of the system is 16221465.4617 kgm²/s

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A 2-kg bowling ball sits on top of a building that is 40 meters tall.

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The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho= rho0[1+α(T−T0)] , where

Stolb23 [73]

Answer:

At 81. 52 Deg C its resistance will be 0.31 Ω.

Explanation:

The resistance of wire = $R_T =\frac{\rho_T \ l}{A}$

Where $R_T$ =Resistance of wire at Temperature T

$\rho_T$ = Resistivity at temperature T $=\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]$

Where $T_0 =20\ Deg\ C , \ \rho_0 = Constant, \alpha =3.9 \times 10^-^3 DegC^-1 \ (Given)$

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& A = Area of cross section of wire

For long and thin wire the resistance & resistivity relation will be as follows

$\frac{R_T_1}{R_T_2}=\frac{\rho_0(1+\alpha \cdot(T_1-2 0 )}{\rho_0(1+\alpha \cdot (T_2 -20 )}$

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2 years ago