Answer:
Both A and B
Explanation:
The interaction of magnetic fields and armature results into a rotational force of the armature hence turning motion. It's important to note that you will always need two magnetic fields in order to experience the force since one magnetic field is at the rotating armature and another at the casing. Considering the arguments of these two technicians, both of them are correct in their arguments.
Answer:
Option A
Explanation:
Mechanical waves requires some medium to travel through. They travel faster in the dense medium as compared to a free medium.
The speed of a mechanical wave is fastest in the solid medium and the slowest in the gaseous medium. Hence, as the wave traverses from gaseous medium to the solid medium, its speed increases.
Thus, option A is correct
The answer to your question is capacitor :)
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:
<u>Frictional Force
</u>
When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:
The centripetal acceleration a_c is computed as
Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one
For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as
The normal force N is equal to the weight of the car, thus
Equating both forces
Simplifying
Substituting the values
