Solution: From the given question, we shall find the vector quantity among the
(A) Time , (B) Velocity, (C) Distance , (D) Speed
Concept: <u>Vector Quantity: </u>All those physical quantities which have magnitude as well as specific directions, are called Vector Quantities.
Here, Time, Distance and Speed have only magnitude but have no directions so they will be scalar quantities.
Now, <u>Velocity:</u> It is defined as the change in displacement per unit time. Since the change in the displacement will be in particular direction only. Hence, velocity will be the vector quantity.
Hence, the option (B) Velocity will be the correct option.
D.to bring light together
Answer:
a) F = 64.30 N, b) θ = 121.4º
Explanation:
Forces are vector quantities so one of the best methods to add them is to decompose each force and add the components
let's use trigonometry
Force F1
sin 170 = F_{1y} / F₁
cos 170 = F₁ₓ / F₁
F_{1y} = F₁ sin 170
F₁ₓ = F₁ cos 170
F_{1y} = 100 sin 170 = 17.36 N
F₁ₓ = 100 cos 170 = -98.48 N
Force F2
sin 30 = F_{2y} / F₂
cos 30 = F₂ₓ / F₂
F_{2y} = F₂ sin 30
F₂ₓ = F₂ cos 30
F_{2y} = 75 sin 30 = 37.5 N
F₂ₓ = 75 cos 30 = 64.95 N
the resultant force is
X axis
Fₓ = F₁ₓ + F₂ₓ
Fₓ = -98.48 +64.95
Fₓ = -33.53 N
Y axis
F_y = F_{1y} + F_{2y}
F_y = 17.36 + 37.5
F_y = 54.86 N
a) the magnitude of the resultant vector
let's use Pythagoras' theorem
F = Ra Fx ^ 2 + Fy²
F = Ra 33.53² + 54.86²
F = 64.30 N
b) the direction of the resultant
let's use trigonometry
tan θ’= F_y / Fₓ
θ'= 
θ'= tan⁻¹ (54.86 / (33.53)
θ’= 58.6º
this angle is in the second quadrant
The angle measured from the positive side of the x-axis is
θ = 180 -θ'
θ = 180- 58.6
θ = 121.4º
-- In combination with 610 Hz, the beat frequency is 4 Hz.
So the unknown frequency is either (610+4) = 614 Hz
or else (610-4) = 606 Hz.
In combination with 605 Hz, the beat frequency will be
either (614-605) = 9 Hz or else (606-605) = 1 Hz.
-- In actuality, when combined with the 605 Hz, the beat
frequency is too high to count accurately. That must be
the 9 Hz rather than the 1 Hz.
So the unknown is (605+9) = 614 Hz.