Answer:
i. 43.5 mH ii. 16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°
Explanation:
i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω
The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.
So, x = 2π(50)L = 100πL Ω = 314.16L Ω
Since the current is the same when the 240 V supply is applied, then
the impedance Z = √(R² + X²) = 240 V/15 A
√(R² + X²) = 16 Ω
8.33² + X² = 16²
69.3889 + X² = 256
X² = 256 - 69.3889
X² = 186.6111
X = √186.6111
X = 13.66 Ω
Since X = 314.16L = 13.66 Ω
L = 13.66/314.16
= 0.0435 H
= 43.5 mH
ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.
So in phasor form Z = (8.33 + j13.66) Ω
iii. The phase difference θ between the current and voltage is
θ = tan⁻¹X/R
= tan⁻¹(314.16L/R)
= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)
= tan⁻¹(13.66/8.33)
= tan⁻¹(1.6406)
= 58.64°
Greater the mass greater is inertia. Greater the inertia greater is the force required to stop motion of an object. Linear motion depends only on mass whereas rotational motion depends on mass, size and shape of an object. So conclusion is that it would be difficult to stop 10 kg mass cuz of greater inertia compared to 1kg mass. Hope it clears your doubt.
Answer:
I am sorry I can't draw graphical ok how to draw the graph where what is your position the displacement of time and work 7 kilometres east in 2 hours and what will happen to the time and 72 in 1 hour what is the displacement you after take the displacement formula that is total time taken divided by the distance travelled ok displacement and distance travelled is different about its terms ok
Answer:
Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m
Explanation:
The attractive force experienced by two mass objects is known as Gravitational force.
The gravitational force is determine by the relation:
....(1)
According to the problem,
Mass of Moon, m₁ = 7.35 x 10²² kg
Mass of Earth, m₂ = 5.97 x 10²⁴ kg
Gravitational force experienced by them, F = 1.98 x 10²⁰ N
Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²
Substitute these values in equation (1).
d = 3.85 x 10⁸ m
