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Murljashka [212]
3 years ago
10

What is the angular momentum L of a rotating wheel with mass m, radius r, moment of inertia I, angular velocity ω, and velocity

v of its outer edge? A) mvr B) mωr C) Iω D) mIω
Physics
2 answers:
Angelina_Jolie [31]3 years ago
6 0
<span>The angular momentum L of a rotating wheel with mass m, radius r, moment of inertia I, angular velocity ω, and velocity v of its outer edge:

</span><span>C) Iω</span>
Mrrafil [7]3 years ago
5 0
The angular momentum L of a rotating wheel with mass m, radius r, moment of inertia I, angular velocity ω, and velocity v of its outer edge is "lω<span>". The correct option among all the options that are given in the question is the first option or option "C". I hope that the answer has helped you.</span>
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A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is
lesantik [10]

Answer:

35870474.30504 m

Explanation:

r = Distance from the surface

T = Time period = 24 h

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m = Mass of the Earth =  5.98 × 10²⁴ kg

Radius of Earth = 6.38\times 10^6\ m

The gravitational force will balance the centripetal force

\dfrac{GMm}{R^2}=m\dfrac{v^2}{R}\\\Rightarrow v=\sqrt{\dfrac{GM}{R}}

T=\dfrac{2\pi r}{v}\\\Rightarrow T=\dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}}

From Kepler's law we have relation

T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m

Distance from the center of the Earth would be

42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}

8 0
3 years ago
Astronaut Sarah leaves Earth in a spaceship at a speed of 0.280c relative to Earth. Sarah's destination is a star-system 12.5 li
Usimov [2.4K]

Answer:

L= 12 light years

Explanation:

for length dilation we use the formula

L=L_0\sqrt{1-\frac{v^2}{c^2} }

now calculating Lo

Lo = 12.5×365×24×3600×3×10^8

= 1.183×10^17 m

now putting the values of v and Lo in the above equation we get

L=1.183\times10^{17}\sqrt{1-\frac{0.28c^2}{c^2} }

= 1.136×10^17 m

L=  = \frac{1.136\times10^{17}}{365\times24\times3600\times3\times10^8}m

so L= 12 light years

8 0
4 years ago
Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe
AlexFokin [52]

Answer:

0.56 atm

Explanation:

First of all, we need to find the number of moles of the gas.

We know that

m = 1.00 g is the mass of the gas

Mm=44.0 g/mol is the molar mass of the carbon dioxide

So, the number of moles of the gas is

n=\frac{m}{M_m}=\frac{1.00 g}{44.0 g/mol}=0.023 mol

Now we can find the pressure of the gas by using the ideal gas equation:

pV=nRT

where

p is the pressure

V=1.00 L = 0.001 m^3 is the volume

n = 0.023 mol is the number of moles

R=8.314 J/mol K is the gas constant

T=25.0^{\circ}+273=298 K is the temperature of the gas

Solving the equation for p, we find

p=\frac{nRT}{V}=\frac{(0.023 mol)(8.314 J/mol K)(298 K)}{0.001 m^3}=5.7 \cdot 10^4 Pa

And since we have

1 atm = 1.01\cdot 10^5 Pa

the pressure in atmospheres is

p=\frac{5.7\cdot 10^4 Pa}{1.01\cdot 10^5 Pa/atm}=0.56 atm

5 0
4 years ago
When buying a car, you tell the salesman that you want to make sure the car can go from 0 miles per hour to 60 miles per hour in
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Acceleration....................................... <span />
4 0
3 years ago
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An example of an indicator species is the sea otter. Without sea otters, sea urchins would overgraze on kelp beds, dramatically
sveticcg [70]

the answer is false

7 0
3 years ago
Read 2 more answers
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