A. Larger. It is larger Bc they r all larger than the other
Answer:
<em>600N.</em>
Explanation:
From the question, we are to calculate the net force acting on the car.
According to Newton's second law of motion:
F = ma
m is the mass of the car
a is the acceleration = change in velocity/Time
a = v-u/t
F = m(v-u)/t
v is the final velocity = 30m/s
u is the initial velocity = 20m/s
t is the time = 5secs
m = 300kg
Get the net force:
Recall that: F = m(v-u)/t
F = 300(30-20)/5
F = 60(30-20)
F = 60(10)
<em>F = 600N</em>
<em>Hence the net force acting on the car is 600N.</em>
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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
Answer:
The smallest part of a millimeter that can be read with a digital caliper with a four digit display is 0.02mm. Thus, it has to be converted to centimetre. So, divide by 10, we then have 0.02/10= *0.002cm* not mm.
