The magnitude of the resultant force is to be 200 N, directed along the positive ݒaxis. Determine the magnitude of the force Fin
d the angle θ.
1 answer:
Answer: F = 776.18N θ = 29.41°
Question:
The magnitude of the resultant force is to be 200 N, directed along the positive y-axis. Determine the magnitude of the force F Find the angle θ.
Attached is the image.
Explanation:
To solve this question we need to resolve the forces to the x and y axis.
For the x axis;
Fcosθ = 700cos15....1
For the y axis;
200 = Fsinθ - 700sin15
Fsinθ = 200 + 700sin15 ...2
Dividing equation 2 by 1
Fsinθ/Fcosθ = (200 + 700sin15)/700cos15
Tanθ = (200 + 700sin15)/700cos15
θ = Taninverse(200 + 700sin15)/700cos15)
θ = 29.41°
From equation 1;
F = 700cos15/cosθ
F = 700cos15/cos29.41
F = 776.18N
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Answer:
magnetic flux ΦB = 0.450324 × weber
current I = 1.02484 A
Explanation:
Given data
length a = 2.2 cm = 0.022 m
width b = 0.80 cm = 0.008 m
Resistance R = 0.40 ohms
current I = 4.7 A
speed v = 3.2 mm/s = 0.0032 m/s
distance r = 1.5 b = 1.5 (0.008) = 0.012
to find out
magnitude of magnetic flux and the current induced
solution
we will find magnitude of magnetic flux thorough this formula that is
ΦB = ( μ I(a) /2 π ) ln [(r + b/2 ) /( r -b/2)]
here μ is 4π × put all value
ΦB = (4π × 4.7 (0.022) /2 π ) ln [(0.012+ 0.008/2 ) /( 0.012 -0.008/2)]
ΦB = 0.450324 × weber
and
current induced is
current = ε / R
current = μ I(a) bv / 2πR [(r² ) - (b/2 )² ]
put all value
current = μ I(a) bv / 2πR [(r² ) - (b/2 )² ]
current = 4π × (4.7) (0.022) (0.008) (0.0032) / 2π(0.40) [(0.012² ) - (0.008/2 )² ]
current = 1.02484 A