Answer:
30°
Explanation:
According to the second law of reflection, it States that the angle of incidence i is equal to the angle of reflection r.
The angle of incidence is known to be the angle between the incident ray and the normal.
The Angle of reflection is the angle between the reflected ray and the normal.
This normal ray is a ray that is perpendicular to the surface.
According to the question, if the beam of light is reflected off the surface and its angle of incidence is 30°, its angle of reflection will also be 30° i.e i=r = 30°
Answer:
(1) The orbits are ellipses, with focal points ƒ1 and ƒ2 for the first planet and ƒ1 and ƒ3 for the second planet. The Sun is placed in focal point ƒ1.
(2) The two shaded sectors A1 and A2 have the same surface area and the time for planet 1 to cover segment A1 is equal to the time to cover segment A2.
(3) The total orbit times for planet 1 and planet 2 have a ratio a13/2 : a23/2
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The two properties of the leaf that allow it to be affected by air resistance include its distance from the ground and its mass.
<h3>What is mass?</h3>
Mass is a scalar quantity that measures the inertia of a given body and/or object, which represent one of the most important characteristics of the matter.
This measurement (mass) may affect the flight of an object because mass inertia is associated with gravity and does not allow its movement.
In conclusion, the two properties of the leaf that allow it to be affected by air resistance include its distance from the ground and its mass.
Learn more about mass and inertia here:
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Answer:
Electric Field a the centre
Explanation:
<u>Given:</u>
Total charge on the semicircle =Q
Radius of the semicircle=R
Let consider a elemental charge on the semicircle at an angle
with the horizontal. Since the The charge distribution is symmetrical about y axis so the there will symmetrical charge on other side. The horizontal component will cancel out each other and vertical component will get added so let dE be the electric Field due to elemental charge
Let
be the charge per unit length such that

Total Electric Field at the centre

integrating 0 to 


So the Electric field at the centre is calculated.