A jet fighter flies from the airbase A 300 km East to the point M. Then 350 km at 30° West of North.
It means : at 60° North of West. So the distance from the final point to the line AM is :
350 · cos 60° = 350 · 0.866 = 303.1 km
Let`s assume that there is a line N on AM.
AN = 125 km and NM = 175 km.
And finally jet fighter flies 150 km North to arrive at airbase B.
NB = 303.1 + 150 = 453.1 km
Then we can use the Pythagorean theorem.
d ( AB ) = √(453.1² + 125²) = √(205,299.61 + 15,625) = 470 km
Also foe a direction: cos α = 125 / 470 = 0.266
α = cos^(-1) 0.266 = 74.6°
90° - 74.6° = 15.4°
Answer: The distance between the airbase A and B is 470 km.
Direction is : 15.4° East from the North.
When a swimmer pushes through water to swim they are propelled forward because of the water resistance against the hand and feet. It's A. The water doesn't automatically push the swimmer forward. It releases a reaction after the swimmer pushes through the water.
Answer: MR²
is the the moment of inertia of a hoop of radius R and mass M with respect to an axis perpendicular to the hoop and passing through its center
Explanation:
Since in the hoop , all mass elements are situated at the same distance from the centre , the following expression for the moment of inertia can be written as follows.
I = ∫ r² dm
= R²∫ dm
MR²
where M is total mass and R is radius of the hoop .
Answer:
Q1: 3.2km
Q2: 4.8K
Explanation:
Q1:
So db is the distance of bird, and dr is the distance of runner
db = 2vr and the distance of bird is going to be 2 times greater than the runner.
formulas: db = 2vr & db = 2dr
- db = 2dr
- L + (L - x) = 2x
- 2L - x = 2x
- 2L = 3x
- x =
L
Insert it in x = L
(2.4km) = 1.6km
Now we use formula db = 2dr
- db = 2L - x
- db = 2(2.4km) - 1.6km
- <u>db = 3.2km</u>
Q2:
Formulas: Vr = L /Δt & Vb = db/Δt
- Vr = L/ Δt ⇒ Δt =
(Km cancel each other)
- Vb = db/Δt ⇒ db = VbΔt
- 13.6km/hr
- <u>4.8km</u>
(hr cancel each other)
Hope it helps you :)
