Newton's three laws of motion can be used to describe the motion of the ice skating.
<h3>Newton's first law of motion</h3>
Newton's first law of motion states that an object at rest or uniform motion in a straight line will continue in that state unless it is acted upon by an external force.
- Based on this law, once the ice skating starts, it will continue endlessly unless external force stops it.
<h3>Newton's second law of motion</h3>
Newton's second law of motion states that the force applied to an object is directly proportional to the product of mass and acceleration of an object.
- Based on this law, the force applied to the ice skating is equal to the product of mass and acceleration of the ice skating.
<h3>Newton's third law of motion</h3>
This law states that action and reaction are equal and opposite.
- Based on this law, the force applied to the ice skating is equal in magnitude to the reaction of ice.
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Answer:132.0285
Explanation: Hope this helps!
Answer: I = 111.69 pA
Explanation: The hall effect is all about the fact that when a semiconductor is placed perpendicularly to a magnetic field, a voltage is generated which could be measured at right angle to the current path. This voltage is known as the hall voltage.
The hall voltage of a semiconductor sensor is given below as
V = I×B/qnd
Where V = hall voltage = 1.5mV =1.5/1000=0.0015V
I = current =?,
n= concentration of charge (electron density) = 5.8×10^20cm^-3 = 5.8×10^20/(100)³ = 5.8×10^14 m^-3
q = magnitude of an electronic charge=1.609×10^-19c
B = strength of magnetic field = 5T
d = thickness of sensor = 0.8mm = 0.0008m
By slotting in the parameters, we have that
0.0015 = I × 5/5.8×10^14 × 1.609×10^-19×0.0008
0.0015 = I×5/7.446×10^-8
I = (0.0015 × 7.446×10^-8)/5
I = 111.69*10^(-12)
I = 111.69 pA
The applicable relationship is N1/N2 = V1/V2, meaning the ratio of primary voltage to secondary voltage is equal to the ratio of primary turns to secondary turns.
Here N1 = 1000, V1 = 250, V2 = 400V and N2 = TBD.
Rewriting the above relationship, N2 = N1 V2/V1 = 1000 x 400/250 = 1600 turns.
The expression for the radius and height of the cone can be obtained from
the property of a function at the maximum point.
- The height of the cone is half the length of the radius of the circular sheet metal.
Reasons:
The part used to form the cone = A sector of a circle
The length of the arc of the sector = The perimeter of the circle formed by the base of the cone.
θ/360·2·π·s = 2·π·r
Where;
s = The radius of he circular sheet metal
h = s² - r²
3·r²·s² - 4·r⁴ = 0
3·r²·s² = 4·r⁴
3·s² = 4·r²
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