A piece of aluminium foil is held between a bar magnet and a paperclip. What will happen to the paper clip?

A toroidal solenoid has an inner radius of 12.0 cm and an outer radius of 15.0 cm . It carries a current of 1.50 A . Part A How

tensa zangetsu [6.8K]

Answer:

The number of turns is $N = 1750 \ turns$

Explanation:

From the question we are told that

The inner radius is $r_i = 12.0 \ cm = 0.12 \ m$

The outer radius is $r_o = 15.0 \ cm = 0.15 \ m$

The current it carries is $I = 1.50 \ A$

The magnetic field is $B = 3.75 mT = 3.75 *10^{-3} \ T$

The distance from the center is $d = 14.0 \ cm = 0.14 \ m$

Generally the number of turns is mathematically represented as

$N = \frac{2 * \pi * d * B}{ \mu_o * r_o }$

Generally $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} \ N/A^2$

So

$N = \frac{2 * 3.142 * 0.14 * 3.75 *10^{-3} }{ 4\pi * 10^{-7} * 0.15 }$

$N = 1750 \ turns$

5 0

3 years ago

Suppose we have a 600 kilogram great "yellow" shark swimming to the right at a speed of 3 meters traveled each second as it trie

sammy [17]

Answer:

2.64 m/s

Explanation:

Given that a 600 kilogram great "yellow" shark swimming to the right at a speed of 3 meters traveled each second as it tries to get lunch. An unsuspecting 100 kilogram blue fin tuna is minding its own business swimming to the left at a speed of 0.5 meters traveled each second. GULP! After the great "yellow" shark "collides" with the blue fin tuna

Momentum = MV

Momentum of the yellow shark before collision = 600 × 3 = 1800 kgm/s

Momentum of the tun final before collision = 100 × 0.5 = 50 kgm/s

Total momentum before collision = 1800 + 50 = 1850 kgm/s

Let's assume that they move together after collision. Then,

1850 = ( 600 + 100 ) V

1850 = 700V

V = 1850 / 700

V = 2.64285 m/s

Therefore, the momentum of the shark after collision is 2.64 m/ s approximately

6 0

3 years ago

A worker assigned to the restoration of the Washington Monument is checking the condition of the stone at the very top of the mo

Svetllana [295]

Answer:

The gravitational potential energy of the nickel at the top of the monument is 8.29 J.

Explanation:

We can find the gravitational potential energy using the following formula.

$GPE=mgh$

Identifying given information.

The nickel has a mass $m=0.005 \,kg$ , and it is a the top of Washington Monument.

The Washington Monument has a height of $h=555 \, ft$ , thus we need to find the equivalence in meters using unit conversion in order to find the gravitational potential energy.

Converting from feet to meters.

Using the conversion factor 1 m = 3.28 ft, we have

$h = 555 \, ft \times \cfrac{1 \, m}{3.28 \, ft}$

That give u s

$h = 169.2 \, m$

Finding Gravitational Potential Energy.

We can replace the height and mass on the formula

$GPE=mgh$

And we get

$GPE=(0.005)(9.8)(169.2) \, J$

$\boxed{GPE=8.29 \,J}$

The gravitational potential energy of the nickel at the top of the monument is 8.29 J.

7 0

3 years ago

Read 2 more answers

Ccording to coulomb's law, which pair of charged particles has the lowest potential energy? according to coulomb's law, which pa

Sladkaya [172]

Coulombs law says that the force between any two charges depends on the amount of charges and distance between them. This force is directly proportional to the magnitude of the two charges and inversely proportional to the distance between them.

$F=k\frac{|q_1| |q_2|}{r^2}$

where $q_1\hspace{1mm}and\hspace{1mm}q_2$ are charges, $r$ is the distance between them and k is the coulomb constant.

case 1:

$q_1=-e\\ q_2=+3e\\ r=100pm\\ \Rightarrow F=k\frac{|-e||3e|}{(100pm)^2}=3ke^2\times10^8$

case 2

$q_1=-e\\ q_2=+2e\\ r=100pm\\ \Rightarrow F=k\frac{|-e||2e|}{(100pm)^2}=2ke^2\times10^8$

case 3:

$q_1=-e\\ q_2=+e\\ r=100pm\\ \Rightarrow F=k\frac{|-e||e|}{(200pm)^2}=0.25ke^2\times10^8$

Comparing the 3 cases:

The maximum potential force according to coulombs law is between -1 charge and +3 charge separated by a distance of 100 pm.

3 0

3 years ago

Read 2 more answers

Without using a micrometer screw gauge, how do I find the average diameter of a long piece of thin wire using a metre rule and a

Mice21 [21]

Answer:

Wind the long piece of thin wire around the uniform glass rod multiple times, find the length of the total diameters using the metre ruler, and divide by the number of times you wound it around the rod.

Explanation:

Since the diameter of one long piece of thin wire is too thin to be measured by a metre ruler, you can wind it multiple times and push it side by side to get a length you can measure.

For example, if you wound it around 20 times and the total length of 20 diameters of the wire side-by-side is 2.0 cm, one winding, which is the diameter would be 2.0cm ÷ 20 = 0.10cm or 1mm.

5 0

2 years ago