Answer:
If I remembered correctly the answer would be Ne and Ar(Neon and Argon
The angle of incidence for a ray of light passing through the center of curvature of a concave mirror is 0°.
The angle of incidence is the angle between the surface's normal and the incident ray. For a concave mirror, the normal of the surface is along the center of the curvature, and a ray of light passed through a center of curvature passes through the normal of the surface.
The ray of light retreats its path making a zero angle of reflection. The law of reflection state that the angle of incidence is equal to the angle of reflection; therefore, the angle of incidence of a concave surface passed through the center of curvature is zero degrees.
Learn more about the angle of incidence here:
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Answer:
if it is a plastic connector it wont work but if there is metal or steel it will work
Explanation:
Answer:
The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.
Explanation:
Given that
q₁ = 5 μ C
q₂ = - 4 μ C
The distance between charges = 50 cm
d= 50 cm
Lets take at distance x from the charge μ C ,the electrical field is zero.
That is why the distance from the charge - 4 μ C = 50 - x cm
We know that ,electric field is given as
Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.
Answer:
vDP = 21.7454 m/s
θ = 200.3693°
Explanation:
Given
vDE = 7.5 m/s
vPE = 20.2 m/s
Required: vDP
Assume that
vDE to be in direction of - j
vPE to be in direction of i
According to relative motion concept the velocity vDP is given by
vDP = vDE - vPE (I)
Substitute in (I) to get that
vDP = - 7.5 j - 20.2 i
The magnitude of vDP is given by
vDP = √((- 7.5)²+(- 20.2)²) m/s = 21.7454 m/s
θ = Arctan (- 7.5/- 20.2) = 20.3693°
θ is in 3rd quadrant so add 180°
θ = 20.3693° + 180° = 200.3693°
