Answer:
F = 102 N
Explanation:
ASSUMING the track is horizontal to avoid gravity potential energy considerations.
The work of the applied force must overcome friction work and increase the kinetic energy of the runner
F(d) = Ff(d) + ½m(v² - u²)
F = Ff + m(v² - u²) / 2d
F = 30.0 + 60.0(8.00² - 2.00²) / (2(25.0))
F = 102 N
Answer: The correct option is A ( horizontally towards the east)
Explanation:
Magnetic field is a region around a magnet or a current- carrying conductor, where a magnetic force is experienced. The magnetic effect of electric current was first discovered in the early 1820 by Oersted. Using a wire that had current flowing through it and a pivoted magnetic needle, he discovered that the direction of deflection depended on the direction of the current and whether the wire was above or below the needle.
From the way the needle turns when current when current carrying wire is held parallel to it, he therefore concluded that:
--> a current has magnetic field all round it,
--> the magnetic field is in a direction perpendicular to the current.
The above discovery was now modified in Fleming's left hand rule which states that when conductor carrying current is placed in a magnetic field, the conductor will experience a force perpendicular to both the field and the flow of current.
Therefore from the question, a vertical wire carrying current in DOWNWARD direction is placed in a HORIZONTAL magnetic field directed to the NORTH. The direction of the force on the wire is to the EAST.
The properti that you didnt share
A :-) for this question , we should apply
F = ma
Given - F = 12 N
a = 0.20 m/s^2
Solution -
F = ma
12 = m x 0.20
m = 12 by 0.20
m = 60 kg
.:. The mass is 60 kg.
Answer:
C. 0.25J
Explanation:
Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;
L is the inductance
I is the current flowing in the inductor
Given parameters
L = 20mH = 20×10^-3H
I = 5A
Required
Energy stored in the magnetic field.
E = 1/2 × 20×10^-3 × 5²
E = 1/2 × 20×10^-3 × 25
E = 10×10^-3 × 25
E = 0.01 × 25
E = 0.25Joules.
Hence the energy stored in the magnetic field of this inductor is 0.25Joules
