Answer:
A) d_o = 20.7 cm
B) h_i = 1.014 m
Explanation:
A) To solve this, we will use the lens equation formula;
1/f = 1/d_o + 1/d_i
Where;
f is focal Length = 20 cm = 0.2
d_o is object distance
d_i is image distance = 6m
1/0.2 = 1/d_o + 1/6
1/d_o = 1/0.2 - 1/6
1/d_o = 4.8333
d_o = 1/4.8333
d_o = 0.207 m
d_o = 20.7 cm
B) to solve this, we will use the magnification equation;
M = h_i/h_o = d_i/d_o
Where;
h_o = 3.5 cm = 0.035 m
d_i = 6 m
d_o = 20.7 cm = 0.207 m
Thus;
h_i = (6/0.207) × 0.035
h_i = 1.014 m
Answer:
Explanation:
Image formation by objective lens ,
f = focal length of objective = .75 cm, ( positive )
object distance u = .81 cm ( negative )
v = image distance
1/v - 1/u = 1/f
1/v + 1/.81 = 1 / .75
1/v = 1/ .75 - 1/.81 = .098765
v = 10.125.cm
Image formation by eye piece,
v = infinity ( for relaxed eye )
f ( eye piece ) = 2.70cm
u = ?
1/v - 1/u = 1/f
0 -1/u = 1/2.7
u = 2.7 cm
Total length between lenses
= 2.7 + 10.125
= 12.825 cm
Total magnification = m₁ x m₂
m₁ is magnification by objective and m₂ is magnification by eye piece
m₁ = v/u = 10.125 / .810 = 12.5
m₂ = 1 + D / f = 1 + 25 / 2.7 = 10.25
Total magnification
= 12.5 x 10.25 = 128.125
Answer:
67000N
Explanation:
We solve for the acceleration using the the 3rd constant-acceleration equation.
(Vx)f² = (Vx)i² + 2ax∆x
We have the displacement to be
∆x = Xf - Xi = 940m
Vx = 70m/s
The acceleration = (70m/s)²/2(940m)
= 4900/1880
= 2.61m/s²
From isaac newton's second law,
51000kg x 2.61m/s²
= 133,000N
The engines thrust is half of this value
Therefore thrust = 67000N or 67kN
Answer:
KE = 1/2mv^2
KE = 1/2(1120)(40^2)
KE = 560(1600)
KE = 896000
Let me know if this helps!
Answer:
Mono-chromatic, Poly-chromatic. When a wave moves from one medium into another at an angle other than 90 degrees the wave will change direction and continue to follow a new straight-line path
<u><em>Thank you</em></u>