A 1.0-m-long copper wire of diameter 0.10 cm carries a current of 50.0 A to the east. Suppose we apply to this wire a magnetic f
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1.
For the first part, we just need to write the equation of the forces along two perpendicular directions.
We have actually only two forces acting on the car, if we want it to go around the track without friction:
- The weight of the car, mg, downward
- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)
By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:
(1)
(2)
where in the second equation, the term represents the centripetal force, with v being the speed of the car and R the radius of the track.
Dividing eq.(2) by eq.(1), we get the following expression:
2.
In this second situation, the cars moves around the track at a speed
This means that the centripetal force term
is now larger than before, and therefore, the horizontal component of the normal reaction, , is no longer enough to keep the car in circular motion.
This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes
And re-arranging for F,
(3)
But from eq.(2) in the previous part we know that
So, susbtituting into eq.(3),
Refer to the diagram shown below.
i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω
Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5 (10
Also,
R₂*i = 9.5 (2)
Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A
From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)
Answer: 0.08 Ω
i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω
Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5 (10
Also,
R₂*i = 9.5 (2)
Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A
From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)
Answer: 0.08 Ω
Answer:
The elastic potential energy is zero.
The net force acting on the spring is zero.
Explanation:
The equilibrium position of a spring is the position that the spring has when its neither compressed nor stretched - it is also called natural length of the spring.
Let's now analyze the different statements:
The spring constant is zero. --> false. The spring constant is never zero.
The elastic potential energy is at a maximum --> false. The elastic potential energy of a spring is given by
where k is the spring constant and x the displacement. Therefore, the elastic potential energy is maximum when x, the displacement, is maximum.
The elastic potential energy is zero. --> true. As we saw from the equation above, the elastic potential energy is zero when the displacement is zero (at the equilibrium position).
The displacement of the spring is at a maxi num --> false, for what we said above
The net force acting on the spring is zero. --> true, as the spring is neither compressed nor stretched