Answer:
Incomplete question
This is the complete question
For a magnetic field strength of 2 T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical nerve that has a diameter of 1.5 mm. Assume that the entire nerve carries a current due to an applied voltage of 100 mV (that of a typical action potential). The resistivity of the nerve is 0.6ohms meter
Explanation:
Given the magnetic field
B=2T
Lenght of rod is 1mm
L=1/1000=0.001m
Diameter of rod=1.5mm
d=1.5/1000=0.0015m
Radius is given as
r=d/2=0.0015/2
r=0.00075m
Area of the circle is πr²
A=π×0.00075²
A=1.77×10^-6m²
Given that the voltage applied is 100mV
V=0.1V
Given that resistive is 0.6 Ωm
We can calculate the resistance of the cylinder by using
R= ρl/A
R=0.6×0.001/1.77×10^-6
R=339.4Ω
Then the current can be calculated, using ohms law
V=iR
i=V/R
i=0.1/339.4
i=2.95×10^-4 A
i=29.5 mA
The force in a magnetic field of a wire is given as
B=μoI/2πR
Where
μo is a constant and its value is
μo=4π×10^-7 Tm/A
Then,
B=4π×10^-7×2.95×10^-4/(2π×0.00075)
B=8.43×10^-8 T
Then, the force is given as
F=iLB
Since B=2T
F=iL(2B)
F=2.95×10^-4×2×8.34×10^-8
F=4.97×10^-11N
Answer:
3675 J
Explanation:
Gravitational Potential Energy = 
× mass × g × height
( g is the gravitation field strength )
Mass = 50 kg
G = 9.8 N/kg ( this is always the same )
Height = 15 m
Gravitational Potential Energy = × 50 ×9.8 × 15
= 3675 J
the equation of the tangent line must be passed on a point A (a,b) and
perpendicular to the radius of the circle. <span>
I will take an example for a clear explanation:
let x² + y² = 4 is the equation of the circle,
its center is C(0,0). And we assume that the tangent line passes to the point
A(2.3).
</span>since the tangent passes to the A(2,3), the line must be perpendicular to the radius of the circle.
<span>Let's find the equation of the line parallel to the radius.</span>
<span>The line passes to the A(2,3) and C (0,0). y= ax+b is the standard form of the equation. AC(-2, -3) is a vector parallel to CM(x, y).</span>
det(AC, CM)= -2y +3x =0, is the equation of the line // to the radius.
let's find the equation of the line perpendicular to this previous line.
let M a point which lies on the line. so MA.AC=0 (scalar product),
it is (2-x, 3-y) . (-2, -3)= -4+4x + -9+3y=4x +3y -13=0 is the equation of tangent
Gravitational potential energy = mass x acceleration due to gravity x height
GPE=mgh
4620=mx9.81x8.4
4620/(9.81x8.4)=m=56.1 kg
Answer:
3.6 KJ
Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
The workdone = the energy.
There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )
P.E = mgh
P.E = 70 × 9.8 × 1.6
P.E = 1097.6 J
P.E = 1.098 KJ
K.E = 1/2mv^2
K.E = 1/2 × 70 × 8.5^2
K.E = 2528.75 J
K.E = 2.529 KJ
The non conservative workdone = K.E + P.E
Work done = 1.098 + 2.529
Work done = 3.63 KJ
Therefore, the non conservative workdone is 3.6 KJ approximately
