Is a electron a positively charged, negatively charged, or a neutral particle

A roller coaster, traveling with an initial speed of 15 meters per second, decelerates uniformly at â7.0 meters per second2 to a

Harrizon [31]

<span>By algebra, d = [(v_f^2) - (v_i^2)]/2a. Thus, d = [(0^2)-(15^2)]/(2*-7) d = [0-(225)]/(-14) d = 225/14 d = 16.0714 m With 2 significant figures in the problem, the car travels 16 meters during deceleration.</span>

8 0

3 years ago

Which of the following information does the electromagnetic spectrum tell us?

Elodia [21]

A: the wavelength of a wave

8 0

2 years ago

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A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

Elden [556K]

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

You own a high speed digital camera that can take a picture every 0.5 seconds. You decide to take a picture every 0.5 seconds of

Vesnalui [34]

-- There is no need to develop the pictures. They are available immediately in a digital camera.

-- There is no change in the teacher from one picture to the next.

-- The distance the watermelon falls from the teacher in each new picture is more in each picture than in the picture before it. (C)

8 0

2 years ago

A skydiver prepares to jump out of a plane. Explain how gravity and air resistance will affect the motion of the skydiver before

ahrayia [7]

Before the skydiver opens the parachute, his velocity would be increasing greatly as much as 9.8 m/s². Opening the parachute would increase the surface area to which air may cause resistance. The skydiver then reaches his terminal velocity.

3 0

2 years ago

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