the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range
1 answer:
Answer:
The actual angle is 30°
Explanation:
<h2>Equation of projectile:</h2><h2>y axis:</h2>the velocity is Zero when the projectile reach in the maximum altitude:
When the time is vo/g the projectile are in the middle of the range.
<h2>x axis:</h2>R=Range
**sin(2A)=2sin(A)cos(A)
<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>Let B the actual angle of projectile
2B=60°
B=30°
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Answer:
Explanation:
The electric flux is defined as the multiple of electric field and the area that the electric field passes through, such that
When calculating the electric flux, the angle between the directions of electric field and the area becomes important, especially if the angle is changing with time.
The above formula can be rewritten as follows
where θ is the angle between the electric field and the area of the loop. Note that, the direction of the area of the loop is perpendicular to the plane of the loop.
If the loop is rotating with constant angular velocity ω, then the angle can be written as follows
At t = 0, cos(0) = 1 and the electric flux through the loop is at its maximum value.
Therefore the electric flux can be written as a function of time