Answer:
from the position of the center of the Sun
Explanation:
As we know that mass of Sun and Jupiter is given as
distance between Sun and Jupiter is given as
now let the position of Sun is origin and position of Jupiter is given at the position same as the distance between them
so we will have
from the position of the center of the Sun
The correct answer is letter D. candela. The unit for measuring the rate at which light energy is radiated from a source is the candela. L<span>umen is the unit for measuring the total amount of visible light emitted by a source. Lux is lumen per square meter. </span>
On a similar problem wherein instead of 480 g, a 650 gram of bar is used:
Angular momentum L = Iω, where
<span>I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore </span>
<span>I = 1/12m*2² = 1/3m kg*m² </span>
<span>The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so </span>
<span>ω = 2π * 2 rev/s = 4π s^(-1) </span>
<span>The angular momentum would therefore be </span>
<span>L = Iω </span>
<span>= 1/3m * 4π </span>
<span>= 4/3πm kg*m²/s, where m is the rod's mass in kg. </span>
<span>The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. </span>
<span>Edit: 650 g = 0.650 kg, so </span>
<span>L = 4/3π(0.650) kg*m²/s </span>
<span>≈ 2.72 kg*m²/s</span>
Answer:
0.0619 m^3
Explanation
number of moles = n = 4.39 mol
pressure = P = 2.25 atm =2.25×1.01×10^5 Pa= 2.27×10^5 Pa
Molar gas constant =R = 8.31 J/(mol K)
Temperature T= 385K
volume of gas = V =?
BY GENERAL GAS LAW WE HAVE
PV = nRT
or V = nRT/P
or V = (4.39×8.31×385)/(2.27×10^5)
V = 0.0618728
V = 0.0619 m^3
<span>The entire time the ball is in the air, its acceleration is 9.8 m/s2 down provided this occurs on the surface of the Earth. Note that the acceleration can be either 9.8 m/s2 or -9.8 m/s2.
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