The displacement of Edward in the westerly direction is determined as 338.32 km.
<h3>What is displacement of Edward?</h3>
The displacement of Edward can be determined from different methods of vector addition. The method applied here is triangular method.
The angle between the 200 km north west and 150 km west = 60 + 90 = 150⁰
The displacement is the side of the triangle facing 150⁰ = R
R² = a² + b² - 2abcosR
R² = 150² + 200² - (2x 150 x 200)xcos(150)
R² = 62,500 - (-51,961.52)
R² = 114,461.52
R = 338.32 km
Learn more about displacement here: brainly.com/question/321442
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Answer:
v = √ 2 G M/ 
Explanation:
To find the escape velocity we can use the concept of mechanical energy, where the initial point is the surface of the earth and the end point is at the maximum distance from the projectile to the Earth.
Initial
Em₀ = K + U₀
Final
= 
The kinetic energy is k = ½ m v²
The gravitational potential energy is U = - G m M / r
r is the distance measured from the center of the Earth
How energy is conserved
Em₀ =
½ mv² - GmM / = -GmM / r
v² = 2 G M (1 / – 1 / r)
v = √ 2GM (1 / – 1 / r)
The escape velocity is that necessary to take the rocket to an infinite distance (r = ∞), whereby 1 /∞ = 0
v = √ 2GM /
This means it has TWO poles
The 2 poles are NORTH and SOUTH
<span>Acceleration is the rate of
change of the velocity of an object that is moving. This value is a result of
all the forces that is acting on an object which is described by Newton's
second law of motion. Calculations of such is straightforward, if we are given
the final velocity, the initial velocity and the total time interval. However, we are not given these values. We are only left by using the kinematic equation expressed as:
d = v0t + at^2/2
We cancel the term with v0 since it is initially at rest,
d = at^2/2
44 = a(6.2)^2/2
a = 2.3 m/s^2
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