The bearing of lines A and B are 16° 10` and 332° 18`, the value of the included angle BOA is:_______
1 answer:
Answer:
D. 43° 52`
Explanation:
A bearing is an angle, measured clockwise from the north direction. When solving a bearing problem, it is good to represent the bearings in the given question with diagram.
The diagrammatically representation of the bearing of lines A and B, 16° 10` and 332° 18` respectively given in the question is shown in the figure attached.
At Point A, we will calculate angle ∠BAO.
Calculating the angle ∠BAO
∠BAO = 90° - 16° 10`
= 73° 50`
At Point B, we will calculate angle ∠ABO.
Calculating the angle ∠ABO
∠ABO = 332° 18` - 270° 0`
= 62° 18`
At Point O, we will calculate the include angle ∠BOA.
Calculating the angle ∠BOA
∠BAO + ∠ABO + ∠BOA = 180° (sum of angles in a triangle)
73° 50` + 62° 18` + ∠BOA = 180°
136° 8` + ∠BOA = 180°
∠BOA = 180° - 136° 8`
∠BOA = 43° 52`
The value of the included angle BOA is 43° 52
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Answer: Answer down below.
Explanation:
The vectors adition we can find the magnitude of the force applied by the other astronaut is 11.25 N in the y direction
Parameters given
- Force of an astronaut Fₓ = 42 N
- Angle θ = 15º
To find
- Force another astronaut
The force is a vector magnitude for which the addition of vectors must be used, a very efficient method to perform this sum is to add the components of each vector and devise constructing the resulting vector using trigonometry and the Pythagorean theorem.
Let's use trigonometry to find the other force
tan θ =
F_ y = Fₓ tan θ
let's calculate
F_y = 42 tan 15
F_y = 11.25 N
Using the summation of vectors we can find the magnitude of the force applied by the other astronaut is 11.25 N in the y direction
Learn more about vector addition here:
Answer:
Explanation:
An inelastic collision is one where 2 masses collide and stick together, moving as a single mass after the collision occurs. When we talk about this type of momentum conservation, the momentum is conserved always, but the kinetic momentum is not (the velocity changes when they collide). Because there is direction involved here, we use vector addition. The picture before the collision has the truck at a mass of 3520 kg moving north at a velocity of 18.5. The truck's momentum, then, is 3520(18.5) = 65100 kgm/s; coming at this truck is a car of mass 1480 kg traveling east at an unknown velocity. The car's momentum, then, is 1480v. The resulting vector (found when you pick up the car vector and stick the initial end of it to the terminal end of the truck's momentum vector) forms the hypotenuse of a right triangle where one leg is 65100 kgm/s, and the other leg is 1480v. Since we already know the final velocity of the 2 masses after the collision, we can use that to find the final momentum, which will serve as the resultant momentum vector in our equation (we'll get there in a sec). The final momentum of this collision is
p = mv and
p = (3520 + 1480)(13.6) so
p = 68000. Final momentum. The equation for this is a take-off of Pythagorean's Theorem and the one used to find the final magnitude of a resultant vector when you first began your vector math in physics. The equation is
which, in words, is
the final momentum after the collision is equal to the square root of the truck's momentum squared plus the car's momentum squared. Filling in:
and
and
and
and
so
v = 13.3 m/s at 72.6°