Answer:
Time of flight A is greatest
Explanation:
Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.
So
H = u₁² sin²θ₁ /2g
H = u₂² sin²θ₂ /2g
H = u₃² sin²θ₃ /2g
On the basis of these equation we can write
u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃
For maximum range we can write
D = u₁² sin2θ₁ /g
1.5 D = u₂² sin2θ₂ / g
2 D =u₃² sin2θ₃ / g
1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁
1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )
u₂ cosθ₂ >u₁ cosθ₁
u₂ sinθ₂ < u₁ sinθ₁
2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g
Time of flight B < Time of flight A
Similarly we can prove
Time of flight C < Time of flight B
Hence Time of flight A is greatest .
Answer:
3.86×10⁶ Newton/coulombs
Explaination:
Applying,
E = F/q....................... Equation 1
Where E = Electric Field, F = Force, q = charge.
From the question,
Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs
Substitute these values into equation 1
E = 5.4×10⁻¹/ -1.4×10⁻⁷
E = -3.86×10⁶ Newtons/coulombs
Hence the magnitude of the electric field created by the
negative test charge is 3.86×10⁶ Newton/coulombs
Answer:
the number of turns in the primary coil is 13000
Explanation:
Given the data in the question;
V₁ = 13000 V
V₂ = 120 V
N₁ = ?
N₂ = 120 turns
the relation between the voltages and the number of turns in the primary and secondary coils can be expressed as;
V₁/V₂ = N₁/N₂
V₁N₂ = V₂N₁
N₁ = V₁N₂ / V₂
so we substitute
N₁ = (13000 V × 120 turns) / 120 V
N₁ = 1560000 V-turns / 120 V
N₁ = 13000 turns
Therefore, the number of turns in the primary coil is 13000
F = kq1q2/r<span>2
Where,
F - Coulomb Force
k - constant value which is equal to </span>8.98 × 10^9<span> newton square metre per square coulomb
q1 and q2 - two electric charges
r - distance.
5.8 * 10^5 = 1.5 * 10^-9 / r^2
</span><span>5.8 * 10^5 r^2 = 1.5*10^-9
</span>r^2 = 0.0000258620
r = 0.0050854694
So the distance is equal to 5.09 x 10^-3
