Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
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
31.3m/s
Explanation:
Given parameters:
Mass of rock = 40kg
Height of cliff = 50m
Unknown:
Speed of rock when it hits ground = ?
Solution:
We are going to use the appropriate motion equation to solve this problem
The rock is falling with the aid of gravitational force. The force is causing it to accelerate with an amount of velocity.
Using;
V² = U² + 2gH
V = unknown velocity
U = initial velocity = O
g = acceleration due to gravity = 9.8m/s²
H = height of fall
since the initial velocity of the bodyg is 0
V² = 2gH
V= √2gH = √2 x 9.8 x 50 = 31.3m/s
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Answer:
The workdone is
Explanation:
From the question we are told that
The potential difference is
Generally the charge on is
Generally the workdone is mathematically represented as
=>
=>