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°
Answer:
A. Zero
Explanation:
Given data,
The charge of the test charge, q = 1 C
The distance the charge moved against the filed of intensity, x = 30 cm
= 0.3 m
The electric field intensity, E = 50 N/C
The energy stored in the charge at 0.3 m is given by the formula,
V = k q/r
Where,
= 9 x 10⁹ Nm²C⁻²
The charge is moved from the potential V₁ to V₂ at 30 cm
Substituting the given values in the above equation
V₁ = 9 x 10⁹ x 30 / 0.3
= 1.5 x 10¹² J
And,
V₂ = 1.5 x 10¹² J
The energy stored in it is,
W = V₂ - V₁
= 0
Hence, the energy stored in the charge is, W = 0
The working equation would be Vf (final velocity) = Vi
(initial velocity) + a (acceleration) t (time). The given data are the initial
velocity (5.0 m/s), acceleration (-2.5 m/s^2, negative since it is said to
decelerate) and the final velocity (0 m/s, since it will put to a stop). The
time would be 2 seconds.
Answer:
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