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Vera_Pavlovna [14]
4 years ago
8

A 0.40-kg object is traveling to the right (in the positive direction) with a speed of 4.0 m/s. After a 0.20 s collision, the ob

ject is traveling to the left at 2.0 m/s. What is the magnitude of the impulse (in N-s) acting on the object
Physics
1 answer:
Sav [38]4 years ago
8 0

Answer:

2.4 N\cdot s.

Explanation:

The equation that models the travel of the object is:

m \cdot v{o} + \Sigma F \cdot \Delta t = m \cdot v_{f}

The impulse is:

\Sigma F \cdot \Delta t = m \cdot (v_{f} - v_{o})

By replacing terms:

\Sigma F \cdot \Delta t = (0.40 kg) \cdot [- 2 \frac{m}{s} - 4 \frac{m}{s} ]\\\Sigma F \cdot \Delta t = - 2.4 N \cdot s

The magnitude of the impulse acting on the object is 2.4 N\cdot s.

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The current and the potential difference in an inductor are in phase. B. The current lags the potential difference by π/2 in an
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The current lags the potential difference by π/2 in an inductor

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6 0
3 years ago
A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly
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Answer:

a. B= 9.45 \times10^{-3} T

b. B= 0.820 T

c. B= 0.0584 T

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

J= \frac {-I_{c}}{\pi(c^{2}-b^{2}  ) }}

J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}

B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\

B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0
3 years ago
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