Answer:
1.4m/s
Explanation:
Average velocity is the total distance covered divided by the total time taken.
Average velocity =
Total time taken = 5s + 6s = 11s
The first distance covered = velocity x time = 1.4 x 5 = 7m
second distance covered = velocity x time = 1.4 x 6 = 8.4m
So;
Average velocity =
= 1.4m/s
Answer:
C2, C1, C4, C5 and C6 are in parallel. Therefore, we use the formula Cp = C1 + C2 + ....
Cp = C2 + C1 + C4 + C5 + C6 = ( 7 * 10 ^-3) + (18 * 10^-6) + (0.8F) + (200 * 10^-3 F) + (750 * 10^-6) = 1.008F
Now, Cp will become one capacitor and it will be aligned with C3, therefore it will now become a circuit in series.
We use the formula: 1/Cs = 1/C1 + 1/C2 + .... + ....1/Cn
Thus,
1/Cs = 1/C3 + 1/Cp
1/Cs = 1/(14 * 10^-3 F) + 1/(1.008F)
Cs = 1.4 * 10 ^-2 or if we do not round too much it will give exactly 0.0138 F
So the answer should be a)
Answer is c, they are equal:
Explanation:
Answer:
1 x 10 -10 whisper at 1m distance.
Explanation:
- Properly fitted ear plugs an reduce noise form 15-30db. Although they are better for low frequency
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