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Greeley [361]
3 years ago
14

Find the velocity, acceleration, and speed of a particle with the given position function. r(t = t2i 6tj 4 ln t k

Physics
1 answer:
artcher [175]3 years ago
6 0
1st derivative gives velocity;
d r(t)/ dt = 2t i + 6 j + 4/t k

2nd derivative gives acceleration;
d^2 r(t)/ dt^2 = 2 i - 4/ t^2

Speed ;
Square root of (4 t^2 + 36 + 16/ t^2)

For a given time, like 2 seconds, t will be 2. And answer of speed will be scalar.
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Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire fe
Aleksandr [31]

Complete question:

Two parallel 3.0-meter long wires conduct current. The current in the top wire is 12.5 A and flows to the right. The top wire feels a repulsive force of 2.4 x 10^-4 N created by the interaction of the 12.5 A current and the magnetic field created by the bottom current (I). Find the magnitude and direction of the bottom current, if the distance between the two wires is 40cm.

Answer:

The bottom current is 12.8 A to the right.

Explanation:

Given;

length of the wires, L = 3.0 m

current in the top wire, I₁ = 12.5 A

repulsive force between the two wires, F = 2.4 x 10⁻⁴ N

distance between the two wires, r = 40 cm = 0.4 m

The repulsive force between the two wires is given by;

F = \frac{\mu_oI_1I_2L}{2\pi r}\\\\I_{2} = \frac{2F\pi r}{\mu_oI_1L}

Where;

I₂ is the bottom current

The direction of the bottom current must be in the same direction as the top current since the force between the two wires is repulsive.

I_{2} = \frac{2F\pi r}{\mu_oI_1L}\\\\I_{2} = \frac{2(2.4*10^{-4})(\pi)(0.4)}{(4\pi*10^{-7})(12.5)(3)}\\\\I_{2} = 12.8 \ A

Therefore, the bottom current is 12.8 A to the right.

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<em><u></u></em>

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