Answer:
Static friction
Explanation:
When there is contact between two surfaces and there is no movement, there is static friction. So, if we have a book that rests on a wooden table, we have an example of Static friction, where this force avoids the movement of the book.
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:
Explanation:
fundamental frequency, f = 250 Hz
Let T be the tension in the string and length of the string is l ans m be the mass of the string initially.
the formula for the frequency is given by
.... (1)
Now the length is doubled ans the tension is four times but the mass remains same.
let the frequency is f'
.... (2)
Divide equation (2) by equation (1)
f' = √2 x f
f' = 1.414 x 250
f' = 353.5 Hz
