The magnitude of the resultant momentum can be calculated using the following rule:
p = sqrt [(px)^2 + (py)^2]
px is given to be 100 kg m/sec
py is given to be 98 kg m/sec
Substitute in the above equation to calculate the magnitude of the resultant momentum as follows:
p = sqrt[(100)^2+(98)^2]
p = 140 kg m/sec
Answer:
186 lbs per man.
Explanation:
If we assume that we have 35 men in the elevator, and the elevator will be overloaded if the total weight is in excess of 6510-lbs, so we just need to take average:
6510 lbs / 35 = 186 lbs per man.
Have a nice day!
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°
Answer:
The frequency of oscillation of the simple pendulum is 0.49 Hz.
Explanation:
Given that,
Mass of the simple pendulum, m = 0.35 kg
Length of the string to which it is attached, l = 1 m
We need to find the frequency of oscillation. The frequency of oscillation of the simple pendulum is given by :
So, the frequency of oscillation of the simple pendulum is 0.49 Hz. Hence, this is the required solution.
