Answer:
The magnitude of the magnetic force on the rod is 0.037 N.
Explanation:
The magnetic force is given by:
Since the charge (q) is:
Where<em> I</em> is the current = 1.40 A, and <em>t</em> the time
And the speed (v):
Where <em>L </em>is the tracks separation = 2.20 cm = 0.022 m
Hence, the magnetic force is:
Where <em>B </em>is the magnetic field = 1.20 T and <em>θ</em> is the angle between the tracks and the magnetic field = 90°
Therefore, the magnitude of the magnetic force on the rod is 0.037 N.
I hope it helps you!
Force applied by the machine to over come resistance
Answer:
Walking while alternating knee lifts with each step.
Standing up and sitting down from a chair without using your hands.
Standing with your weight on one leg and raising the other leg to the side or behind you.
Explanation:
Those are the ones that I could best come up with.
Answer:
v = 14 m/s
= 31.3 mph
The answer would be the same if the mass of the car were 2000 kg
Explanation:
Let V be the final velocity of the car after skidding, and v be the initial velocity of the car. Let a be the acceleration of the car and Δx be the distance the car travels after applying brakes (length of the skid marks). Let Fk be the force of friction between the tyres and the road. Let N be the normal force exerted on the car and μ be the co efficient of kinetic friction.
V^2 = v^2 + 2×a×Δx
Now V, the final velocity is zero as the car stops
0 = v^2 + 2×a×Δx
v^2 = -2×a×Δx
v =√-2×a×Δx .....*
Now applying Newton's Second Law
Fnet = m×a
-Fk = m×a
-μ×N = m×a
-μ×m×g = m×a (The mass cancels out)
a = -μ×g
Substituting the value of a back to equation *
v = √-2×(-μ×g)×Δx
v = √-2×(-0.5×9.8)×20
v = 14 m/s
Therefore the speed the car was travelling with v = 14 m/s
which is equal to 31.3 mph
Now if you were to change the mass of the car to 2000 kg the value for v would still be the same. As it is seen above mass cancels out so it does not influence or affect the value of the velocity obtained.
