Answer:
Explanation:
L 1: front radius 950 mm, rear radius 2700 mm, refractive index 1.528;
We shall use lens maker's formula , that is
1/f = (μ-1) ( 1/R₁ - 1/R₂) , μ is refractive index of the lens , R₁ and R₂ are radius of curvature of front and rear curved surface.
1/f₁ = (1.528-1)( 1/950 + 1/2700)
f₁ = 1331 mm
L2: front radius 535 mm, rear radius 500 mm, refractive index 1.550.
1/f₂ = (1.550-1)( 1/535 + 1/500)
f₂ = 470 mm
largest angular magnification possible
= f₁ /f₂
= 1331 / 470
= 2.83 ( approx )
Length between two lenses
=1331 +470
= 1801 mm
= 1.8 m Ans
The answer is that it is true
Answer:
The mass of the heaviest box you will be able to move with this applied force = 61.4 kg
Explanation:
From the diagram attached, the forces acting on the box include the weight of the box, applied force on the box, normal reaction of the surface on the box and the Frictional force in the opposite direction to the applied force.
For the box to be able to move, the applied force must have a horizontal component that at least matches the Frictional force between the box and the surface. This is the force balance in the horizontal direction.
Resolving the applied force into horizontal and vertical components,
Fₓ = 750 cos 25° = 679.73 N
Fᵧ = 750 sin 25° = 316.96 N
Doing a force balance in the vertical axis,
N = (mg + 316.96)
Frictional force = μN = μ (mg + 316.96)
μ = 0.74, g = 9.8 m/s²
Frictional force = Fᵧ
μ (mg + 316.96) = 679.73
0.74(9.8m + 316.96) = 679.73
7.252m + 234.5504 = 679.73
7.252m = 679.73 - 234.5504 = 445.1796
m = (445.1796/7.252)
m = 61.4 kg
Hope this Helps!!!
A) -3.75 meters/second
A=(20^2-80^2)/(2x800)
=(400-6400)/1600
=-6000/1600
=-3.75
B) 16 seconds
t=(20-80)/-3.75
=-60/-3.75
=16
The ball should take twice as long to return to its original position as it took to reach its maximum height, so it should return to its original position at
.