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svetoff [14.1K]

3 years ago

9

You measure the magnetic field of a long, current-carrying wire to be B is equal to start fraction 50 times mu naught over pi en

d fraction at a distance of r = 7.5 cm. How much current is running through the wire?

1 answer:

FromTheMoon [43]3 years ago

5 0

Answer:

I = 1.875 A

Explanation:

For this exercise we use Ampere's law

∫ B . ds = μ₀ I

We use a circular path around the wire whereby B and ds are parallel, whereby the dot product is reduced to the algebraic product

ds = 2π dr

B (2πr) = μ₀ I

I = B 2π R /μ₀

r= 7.5 cm = 0.075 m

calculate

I = (50 μ₀ /π) 2π 0.075 /μ₀

I = 1.875 A

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A 0.48 kg pool cue moving at 3.39 m/s hits the 0.22 kg cue ball that was a rest and the pool cue stops after the impact. What is

Grace [21]

Answer:

7.39 m/s is the velocity of the cue ball after impact.

Explanation:

Given that,

$\text { Mass of the pool cue is } 0.48 \mathrm{kg} .\left(\mathrm{m}_{\mathrm{i}}\right)$

$\text { Moving at velocity } 3.39 \mathrm{m} / \mathrm{s} .(\mathrm{v_i})$

$\text { Mass of the cue ball that } 0.48 \text { mass ball hits is } 0.22 \mathrm{kg} .\left(\mathrm{m}_{\mathrm{f}}\right)$

$\text { We need to find the velocity ( } \mathrm{v}_{\mathrm{f}} \text { ) of the cue ball of mass } 0.22 \mathrm{kg} \text { . }$

We know that,

If two objects collide then the “total momentum” before is equal to the “total momentum” after collision.

$\mathrm{m}_{\mathrm{i}} \mathrm{v}_{\mathrm{i}}=\mathrm{m}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}$

Substitute the given values in the formula.

$0.48 \times 3.39=0.22 \times \mathrm{V}_{\mathrm{f}}$

$\frac{1.6272}{0.22}=\mathrm{V}_{\mathrm{f}}$

$\mathrm{v}_{\mathrm{f}}=7.39 \mathrm{m} / \mathrm{s}$

3 0

4 years ago

A person in bare feet is standing under a tree during a thunderstorm, seeking shelter from the rain. A lightning strike hits the

Anni [7]

Answer:

$I=38.181\ A$ is the current through the body of the man.

$E=34.5\ J$ energy dissipated.

Explanation:

Given:

time for which the current lasted, $t=43\times 10^{-6}\ s$

potential difference between the feet, $V=21000\ V$

resistance between the feet, $R=550\ \Omega$

<u>Now, from the Ohm's law we have:</u>

$I=\frac{V}{R}$

$I=\frac{21000}{550}$

$I=38.181\ A$ is the current through the body of the man.

<u>Energy dissipated in the body:</u>

$E=I^2.R.t$

$E=38.181^2\times 550\times 43\times 10^{-6}$

$E=34.5\ J$

5 0

4 years ago

A rabbit with a mass of 2.0 kg accelerates at 2.0 meters per second squared. Find the net force of the rabbit

scZoUnD [109]

Answer:

4N

Explanation:

Force = mass x acceleration

Given

Mass = 2.0kg

Acceleration = 2.0m/s^2

Force = 2.0 x 2.0

= 4N

5 0

4 years ago

Read 2 more answers

How much heat energy is produced by 0.5 Wh of electrical energy ?

babymother [125]

1.7 Btu

1 watt = 3.41214 Btu/h

1watt * 1h = 3.41214 Btu/h * h

1 = 3.41214 Btu/ (watt*h)/

0.5 watt * h = 0.5 watt*h * 3.41214 Btu/(watt*h) = 1.706 Btu

6 0

3 years ago

Read 2 more answers

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

miss Akunina [59]

Formula of the gravitational force between two particles:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67 \cdot 10^{-11} Nm^2 kg^{-2}$ is the gravitational constant

m1 and m2 are the masses of the two particles

r is their distance

(a) particle A

The gravitational force exerted by particle B on particle A is

$F_B=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=5.14 \cdot 10^{-5} N$ to the right

The gravitational force exerted by particle C on particle A is

$F_C=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=5.6 \cdot 10^{-6} N$ to the right

So the net gravitational force on particle A is

$F_A = F_B + F_C =5.14 \cdot 10^{-5} N+5.6 \cdot 10^{-6} N=5.7 \cdot 10^{-5} N$ to the right

(b) Particle B

The gravitational force exerted by particle A on particle B is

$F_A=G\frac{m_A m_B}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(567 kg)}{(0.500 m)^2}=-5.14 \cdot 10^{-5} N$ to the left

The gravitational force exerted by particle C on particle B is

$F_C=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=8.41 \cdot 10^{-5} N$ to the right

So the net gravitational force on particle B is

$F_B = F_A + F_C =-5.14 \cdot 10^{-5} N+8.41 \cdot 10^{-5} N=3.27 \cdot 10^{-5} N$ to the right

(c) Particle C

The gravitational force exerted by particle A on particle C is

$F_A=G\frac{m_A m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(340 kg)(139 kg)}{(0.500 m+0.250m)^2}=-5.6 \cdot 10^{-6} N$ to the left

The gravitational force exerted by particle B on particle C is

$F_B=G\frac{m_B m_C}{r^2}=(6.67 \cdot 10^{-11}) \frac{(567 kg)(139 kg)}{(0.250 m)^2}=-8.41 \cdot 10^{-5} N$ to the left

So the net gravitational force on particle C is

$F_C = F_B + F_A =-8.41 \cdot 10^{-5} N-5.6 \cdot 10^{-6} N=-8.97 \cdot 10^{-5} N$ to the left

3 0

3 years ago