Answer:
805.48N/m
Explanation:
According to Hookes law
F = Ke
F is the force = mg
F = 2.4×9.8 = 23.52N
e is the extension = 2.92cm = 0.0292m
Force constant K = F/e
K = 23.52/0.0292
K = 805.48N/m
Hence the force constant of the spring is 805.48N/m
Answer:
T=Lnsin
Please check the attached
Explanation:
The torque can simply be calculated by multiplying the length of the rod by the perpendicular force n as shown in the attached figure.
Note that sin90=1
T=Lsin(nsin90)
T=Lsinxn
T=Lnsin
Locate the mode of 12, 3, 5, 17, 3, 18, 5, 11, 11, 15, 3, 9, 3.
zimovet [89]
Answer:
Mode = 3 because it is listed 4 times
Given Information:
Length of wire = 132 cm = 1.32 m
Magnetic field = B = 1 T
Current = 2.2 A
Required Information:
(a) Torque = τ = ?
(b) Number of turns = N = ?
Answer:
(a) Torque = 0.305 N.m
(b) Number of turns = 1
Explanation:
(a) The current carrying circular loop of wire will experience a torque given by
τ = NIABsin(θ) eq. 1
Where N is the number of turns, I is the current in circular loop, A is the area of circular loop, B is the magnetic field and θ is angle between B and circular loop.
We know that area of circular loop is given by
A = πr²
where radius can be written as
r = L/2πN
So the area becomes
A = π(L/2πN)²
A = πL²/4π²N²
A = L²/4πN²
Substitute A into eq. 1
τ = NI(L²/4πN²)Bsin(θ)
τ = IL²Bsin(θ)/4πN
The maximum toque occurs when θ is 90°
τ = IL²Bsin(90)/4πN
τ = IL²B/4πN
torque will be maximum for N = 1
τ = (2.2*1.32²*1)/4π*1
τ = 0.305 N.m
(b) The required number of turns for maximum torque is
N = IL²B/4πτ
N = 2.2*1.32²*1)/4π*0.305
N = 1 turn
