Answer:
1.549 m
Explanation:
Given:
The radius of the circular board, r = 2 m
The probability of hitting the red is given as 0.6
Now, this probability of hitting the red can be conclude as
0.6 = (Area of red)/ (Total area of the board)
Total area of the board = πr² = π × 2²
let the radius of the red area be R
thus, area of red circle, = πR²
on substituting the value of the area, we have
0.6 = (πR²)/ (π × 2²)
or
R² = 2.4
or
R = 1.549 m
Thus, the radius of the red circle is 1.549 m
Answer:
the correct one is the first, the refractive index of the two materials must be the same
Explanation:
When a beam of light passes through two materials, it must comply with the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
where n₁ and n₂ are the refractive indices of each medium.
In this case, it indicates that the light does not change direction, so the input and output angle of the interface must be the same,
θ₁ = θ₂ = θ
substituting
n₁ = n₂
therefore the refractive index of the two materials must be the same
When reviewing the answers, the correct one is the first
The orbiting speed of the satellite orbiting around the planet Glob is 60.8m/s.
To find the answer, we need to know about the orbital velocity a satellite.
<h3>What's the expression of orbital velocity of a satellite?</h3>
- Mathematically, orbital velocity= √(GM/r)
- G= gravitational constant= 6.67×10^(-11) Nm²/kg², M = mass of sun , r= radius of orbit
<h3>What's the orbital velocity of the satellite in a circular orbit with a radius of 1.45×10⁵ m around the planet Glob of mass 7.88×10¹⁸ kg?</h3>
- Here, M= 7.88×10¹⁸ kg, r= 1.45×10⁵ m
- Orbital velocity of the orbiting satellite = √(6.67×10^(-11)×7.88×10¹⁸/1.45×10⁵)
= 60.8m/s
Thus, we can conclude that the speed of the satellite orbiting the planet Glob is 60.8m/s.
Learn more about the orbital velocity here:
brainly.com/question/22247460
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