Answer:1). Distance of far point x=0.9m
Therefore, since the image is virtual
-f=-x = -0.9m
Power of the concave lenses = 1/f = 1/-0.9
= -1.11D
2 ) near point is 21cm = 0.21m
Power = 4-1/near point
= 4/0.21
= 14.2D.
Answer:
A.
Explanation:
X represents the transmitting power Modulates (amplitude or frequency am/fm), amplifies the signal, transmitting it out at whatever direction the antenna is set up for.
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
Answer should be <span>t=<span>−<span><span>1<span> or </span></span>t</span></span></span>=<span>5</span>