Answer:
No the angular and linear size will not be same for different objects bearing the same angular size and is dependent on the point of observation. The greater the distance between the observer and the object, the greater the difference between the angular and linear size.
Explanation:
The angular size is the ratio of two lengths namely
Here if the distance from the object is unity, the angular and linear size will be similar. However if the distance of observation is very large, the angular size for large bodies with large linear size will be very small.
An example in this regard is Sun.
The linear size of Sun, (the diameter) is 1.3927 million km. Which is very large. However as it is very far from earth, 147.44 million km, the angular size is very small.
it is given as
Now the same angular size can be of a tennis ball having a diameter of 10 cm , placed at around 10.6 m away.
The image of the triangle is to be formed by rotating ΔXYZ 180 degrees about the (2, -3) as shown in the graph.
<h3>What is Geometry?</h3>It deals with the size of geometry, region, and density of the different forms both 2D and 3D.
Triangle XYZ has vertices X(0, 2), Y(4, 4), and Z(3, –1).
If the triangle is ΔXYZ. Then the image of the triangle is to be formed by rotating ΔXYZ 180 degrees about the (2, -3) as shown in the graph.
More about the geometry link is given below.
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Answer:
The magnitude of the free-fall acceleration at the orbit of the Moon is (
, where
).
Explanation:
According to the Newton's Law of Gravitation, free fall acceleration (), in meters per square second, is directly proportional to the mass of the Earth (
), in kilograms, and inversely proportional to the distance from the center of the Earth (
), in meters:
(1)
Where:
- Gravitational constant, in cubic meters per kilogram-square second.
- Mass of the Earth, in kilograms.
- Distance from the center of the Earth, in meters.
If we know that ,
and
, then the free-fall acceleration at the orbit of the Moon is: