Answer:
The field strength needed is 0.625 T
Explanation:
Given;
angular frequency, ω = 400 rpm = (2π /60) x (400) = 41.893 rad/s
area of the rectangular coil, A = L x B = 0.0611 x 0.05 = 0.003055 m²
number of tuns of the coil, N = 300 turns
peak emf = 24 V
The peak emf is given by;
emf₀ = NABω
B = (emf₀ ) / (NA ω)
B = (24) / (300 x 0.003055 x 41.893)
B = 0.625 T
Therefore, the field strength needed is 0.625 T
Answer:
#WeirdestQuestionOfAllTime
Explanation:
Answer: (a) 36.18mm
(b) 23.52
Explanation: see attachment
Answer:
diameter is 14 mm
Explanation:
given data
power = 15 kW
rotation N = 1750 rpm
factor of safety = 3
to find out
minimum diameter
solution
we will apply here power formula to find T that is
power = 2π×N×T / 60 .................1
put here value
15 ×
= 2π×1750×T / 60
so
T = 81.84 Nm
and
torsion = T / Z ..........2
here Z is section modulus i.e = πd³/ 16
so from equation 2
torsion = 81.84 / πd³/ 16
so torsion = 416.75 / / d³ .................3
so from shear stress theory
torsion = σy / factor of safety
so here σy = 530 for 1020 steel
so
torsion = σy / factor of safety
416.75 / d³ = 530 × 
/ 3
so d = 0.0133 m
so diameter is 14 mm
Answer:
Hydrostatic force = 41168 N
Explanation:
Complete question
A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water so that the top is 4 ft below the surface. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)
Let "x" be the side length submerged in water.
Then
w(x)/base = (4+3-x)/altitude
w(x)/5 = (4+3-x)/3
w(x) = 5* (7-x)/3
Hydrostatic force = 62.5 integration of x * 4 * (10-x)/3 with limits from 4 to 7
HF = integration of 40x - 4x^2/3
HF = 20x^2 - 4x^3/9 with limit 4 to 7
HF = (20*7^2 - 4*7^(3/9))- (20*4^2 - 4*4^(3/9))
HF = 658.69 N *62.5 = 41168 N
