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Marizza181 [45]
3 years ago
8

What is the maximum value of the magnetic field at a distance2.5m from a 100-W light bulb?

Physics
1 answer:
MA_775_DIABLO [31]3 years ago
3 0

To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

I = \frac{P}{A}

I = \frac{P}{4\pi r^2}

Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that

I = \frac{100}{4\pi (2.5)^2}

I = 1.2738W/m^2

The relation between intensity I and E_{max}

I = \frac{E_max^2}{2\mu_0 c}

Here,

\mu_0 = Permeability constant

c = Speed of light

Rearranging for the Maximum Energy and substituting we have then,

E_{max}^2 = 2I\mu_0 c

E_{max}=\sqrt{2I\mu_0 c }

E_{max} = 2(1.2738)(4\pi*10^{-7})(3*10^8)

E_{max} = 30.982 V/m

Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,

B_{max} = \frac{E_{max}}{c}

B_{max} = \frac{30.982 V /m}{3*10^8}

B_{max} = 1.03275 *10{-7} T

Therefore the maximum value of the magnetic field is B_{max} = 1.03275 *10{-7} T

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A driver with a 0.80-s reaction time applies the brakes, causing the car to have acceleration opposite the direction of motion.
jeka94

Answer:

a) During the reaction time, the car travels 21 m

b) After applying the brake, the car travels 48 m before coming to stop

Explanation:

The equation for the position of a straight movement with variable speed is as follows:

x = x0 + v0 t + 1/2 a t²

where

x: position at time t

v0: initial speed

a: acceleration

t: time

When the speed is constant (as before applying the brake), the equation would be:

x = x0 + v t

a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:

x = 0m + 26 m/s * 0.80 s = <u>21 m  </u>

b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):

v = v0 + a* t

0 = 26 m/s + (-7.0 m/s²) * t

-26 m/s / - 7.0 m/s² = t

t = 3.7 s

With this time, we can calculate how far the car traveled during the deacceleration.

x = x0 +v0 t + 1/2 a t²

x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>

4 0
3 years ago
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