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
Well idk if this helps but the formula to solve acceleration is
a=F/m=(100kg)=1.0m/s 2
Answer
given,
mass of the piano = 170 kg
angle of the inclination = 20°
moves with constant velocity hence acceleration = 0 m/s²
neglecting friction
so, force required to pull the piano
F = m g sin θ
F = 170 × 9.81 × sin 20°
F = 570.39 N
so, force required by the man to push the piano is F = 570.39 N
Interval training is simply alternating short bursts (about 30 seconds) of intense activity with longer intervals (about 1 to 2 minutes) of less intense activity. For instance, if your exercise is walking and you're in good shape, you might add short bursts of jogging into your regular brisk walks.
Answer:
4 A
Explanation:
The relationship between current, voltage and resistance in a circuit is given by Ohm's law:

where
V is the voltage
R is the resistance
I is the current
The equation can also be rewritten as

from which we see that the current is inversely proportional to the resistance, R.
In this problem, the initial current is I = 8 A. Then the resistance is doubled:
R ' = 2R
So the new current is

so the current is halved.