That's 105 km that he flew, or 65.2 miles ! I'm absolutely positive
that the crow must have landed and gotten some rest when you
weren't looking. But that had no effect on his displacement when
he got where he was going, so we can continue to solve the problem:
The displacement is the distance and direction from the place
where the crow took off to the place where he landed.
-- It's distance is the hypotenuse of the right triangle whose legs
are 60 km and 45 km.
D² = (60 km)² + (45 km)²
= 3,600 km² + 2,025 km² = 5,625 km²
D = √(5625 km²) = 75 km .
-- It's direction is the angle whose tangent is (45 S / 60 W).
tan⁻¹ (45/60) = tan⁻¹ (0.75) = 36.9° south of west
= 53.1° west of south.
= not exactly southwest but close.
Answer:
None
Explanation:
An scale is the factor by which actual features on ground are enlarged or reduced for representing on a plane. There are different kinds of scales:
- Verbal scale use of words to represent scale information on the map. The distance or linear units are used for depicting this scale on the map. For example: 1 inch = 1 Kilo meter.
- Fractional scale uses the numbers or values for showing the scale instead of words. As the name says, it is represented using a fraction or ratio. Example: 1: 10,000 or 1/10,000
- In large scale more details are shown in a map, however, less area coverage will be shown in a single map as the scale is large and more details are given. Example: 1:500
- Small scale is exactly opposite to the large scale, less details are shown as magnification is not enough, however a large amount of area can be shown in a single map. Example: 1:25,000
- A graphic scale is a bar that has been calibrated to show map distances. On maps that have been reduced or enlarged the original ratio and written scales are incorrect, since the relationship between map distance and real world distance has been altered, graphic scale is enlarged or reduced to the same extent as the map, this makes it the right option.
I hope you find this information useful and interesting! Good luck!
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
