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jekas
1 month ago
12

Point m is located a distance 2d from the midpoint between the two wires. find the magnitude of the magnetic field b1m created a

t point m by wire 1.

Physics
1 answer:
Sav [3.1K]1 month ago
3 0

Reminder: The illustration mentioned in the inquiry is included here as a file.

Response:

The intensity of the magnetic field is B = \frac{0.071 \mu I}{d}

Clarification:

The magnetic field can be calculated using the following equation:

B = \frac{\mu I}{2\pi R}...............(1)

Where I refers to the current flowing through the wires

To find the distance R from point 1 to m, apply the Pythagorean theorem

R = \sqrt{d^{2} + (2d)^{2} }

R = \sqrt{5d^{2} } \\R = d\sqrt{5}

Inserting R into equation (1)

B = \frac{\mu I}{2\pi d\sqrt{5} }

B = \frac{0.071 \mu I}{d}

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kicyunya [3294]

Answer:

\frac{V_{big}}{V_{small}} = n^{2/3}

Explanation:

Let the charge on the large droplet be denoted as Q.

When the radius of the droplet is R, the electric potential for the larger droplet can be expressed as:

V_{big} = \frac{KQ}{R}

If it splits into n identical droplets, let the charge of each be "q" and their radius be "r".

Applying volume conservation gives us:

\frac{4}{3}\pi R^3 = n(\frac{4}{3}\pi r^3)

r = \frac{R}{n^{1/3}}

Now, the potential for the smaller droplets is given as:

V_{small} = \frac{kq}{r}

V_{small} = \frac{K(Q/n)}{\frac{R}{n^{1/3}}}

V_{small} = \frac{1}{n^{2/3}}\frac{KQ}{R}

\frac{V_{big}}{V_{small}} = n^{2/3}

7 0
2 months ago
A bus slows down uniformly from 75.0 km/h to 0 km/h in 21 s. How far does it travel before stopping?
Yuliya22 [3333]

1 hour = 3,600 seconds
1 km = 1,000 meters

75 km/hour = (75,000/3,600) m/s = 20-5/6 m/s

The mean speed during the deceleration is

                                   (1/2)(20-5/6 + 0) = 10-5/12 m/s.

Traveling at this average speed for 21 seconds,
the bus covers

                        (10-5/12) × (21) = 218.75 meters.

7 0
3 months ago
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Keith_Richards [3271]

Answer:

12.1 seconds

Explanation:

t = time duration

u = initial speed

v = final speed = 0

s = distance = 120 m

a = lunar gravity acceleration = 1.67 m/s²

Motion equation

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -1.67\times 120-0^2\\\Rightarrow u=\sqrt{2\times 1.67\times 120}\\\Rightarrow u=20.02\ m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-20.2}{-1.67}\\\Rightarrow t=12.1\ s

The rock takes 12.1 seconds to reach the bottom of the crater.

5 0
2 months ago
A student solving a physics problem for the range of a projectile has obtained the expression r= v20sin(2θ)g where v0=37.2meter/
ValentinkaMS [3465]

The formula for range is:

R = \frac{v_o^2 sin2\theta}{g}

Given values are:

v_0=37.2m/s

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g=9.80m/s^2

Using the equation above,

R = \frac{37.2^2 sin2*14.1}{9.80}

The calculated range is 66.7 meters.

Therefore, the range is approximately 66.1 meters.

5 0
3 months ago
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A bathroom scale operates under gravitational influence. Typically, a reading is captured when your body applies force onto the scale. Yet in this scenario, as both you and the scale move downwards, your body ceases to press against the scale. Consequently, the result is:

<span>The scale reading will instantly drop to zero</span>

7 0
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