The luminous star Alnilam in the Orion belt is 1,340 light-years away from Earth. Use the conversion factor 1 parsec = 3.262 lig
ht-years to find the closest conversion of this distance from light-years to parsecs (pc).
2 answers:
Answer:
The distance to the star Alnilam in the Orion belt is 410.8 Pc
Explanation:
The angle due to the change in position of a nearby object against the background stars is known as parallax. This angle is gotten when the position of the object is measured in January and then in July according to the configuration of the Earth with respect to the Sun in those months (see the image below).
The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth(as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km) is defined as 1 astronomical unit:
Where d is the distance to the star.
(1)
Equation (1) can be rewritten in terms of d:
(2)
Equation (2) represents the distance in a unit known as parsec (pc), which is the distance of a star with a parallax angle of 1 arc second.
For the case of the star Alnilam in the Orion belt ( 1340 light-years):
Since 1 parsec is equivalent to 3.262 light-years, therefore:
⇒
Notice how both , Light years and parsecs are units for distance.
Key terms:
Parsec: Parallax of arc second.
Light-year: Distance traveled by light in one year.
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Answer:
N₂ = 19 turns
Explanation:
A transform is a system with two different windings where the variation of the magnetic beam is the same, if there are no losses in the system we can use Faraday's law
V₁ = -N₁
v₂ = - N₂ \frac{d \Phi_B }{dt}
in this case we look for the number of turns in the second winding
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calculate us
N₂ = 360 6.30/ 120
N₂ = 18.9 turn
The number of turns must be an integer
N₂ = 19 turns
Answer:
Explanation:
Let the minimum coefficient of static friction be .
Given:
Mass of the cylinder =
Radius of the cylinder =
Length of the cylinder =
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Initial velocity of the cylinder (Released from rest) = 0
Since, the cylinder is translating and rolling down the incline, it has both translational and rotational motion. So, we need to consider the effect of moment of Inertia also.
We know that, for a rolling object, torque acting on it is given as the product of moment of inertia and its angular acceleration. So,
Now, angular acceleration is given as:
Also, moment of inertia for a cylinder is given as:
Therefore, the torque acting on the cylinder can be rewritten as:
Consider the free body diagram of the cylinder on the incline. The forces acting along the incline are . The net force acting along the incline is given as:
Now, consider the forces acting perpendicular to the incline. As there is no motion in the perpendicular direction, net force is zero.
So,
Plugging in in equation (2), we get
Now, as per Newton's second law,
Now, torque acting on the cylinder is provided by the frictional force and is given as the product of frictional force and radius of the cylinder.
Therefore, the minimum coefficient of static friction needed for the cylinder to roll down without slipping is given as:
