Answer: 1 dioptre of power of a lens is defined as the unit of measuring the power of optical lens Or curved mirror which is equal to the reciprocal of the focal length. The focal length is measured in the meter. 1 dioptre is equal to 1/m where m is the focal length. Basically dioptre is the SI unit of optical power of the lens.
Answer:
Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.
Explanation:
The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.
When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.
The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.
The velocity of the orbit is given by the relation,
Where
V - velocity of the orbit at a height h from the surface
R - Radius of the second object
G - Gravitational constant
h - height from the surface
The body will be in orbital motion when its kinetic motion is balanced by gravitational force.
Hence, the orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.
Answer:
1. telescope
2.
f- focal length
f- focal length r- the radius of curvature of the mirror
p-the distance of the object from the vertex of the mirror
l-the distance of the figure from the vertex of the mirror
Answer:
Acceleration = 5.77 m/s²
Distance cover in 13 seconds = 487.56 meter
Explanation:
Given:
Final velocity of mobile device = 75 m/s
initial velocity of mobile device = 0 m/s
Time taken = 13 seconds
Find:
Acceleration
Distance cover in 13 seconds
Computation:
v = u + at
75 = 0 + (a)(13)
13a = 75
a = 5.77
Acceleration = 5.77 m/s²
s = ut + (1/2)(a)(t²)
s = (0)(t) + (1/2)(5.77)(13²)
Distance cover in 13 seconds = 487.56 meter
Answer:
Time of flight A is greatest
Explanation:
Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.
So
H = u₁² sin²θ₁ /2g
H = u₂² sin²θ₂ /2g
H = u₃² sin²θ₃ /2g
On the basis of these equation we can write
u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃
For maximum range we can write
D = u₁² sin2θ₁ /g
1.5 D = u₂² sin2θ₂ / g
2 D =u₃² sin2θ₃ / g
1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁
1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )
u₂ cosθ₂ >u₁ cosθ₁
u₂ sinθ₂ < u₁ sinθ₁
2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g
Time of flight B < Time of flight A
Similarly we can prove
Time of flight C < Time of flight B
Hence Time of flight A is greatest .