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3 years ago

basic source of magnetism is a) charged particles alone b)Movement of charged particles c) Magnetic dipoles d)magnetic domains

Physics

2 answers:

DIA [1.3K]3 years ago

7 0

Answer:

C . Magnetic dipoles is the correct

Dima020 [189]3 years ago

6 0

Answer:

b). movement of charged particles.

Explanation:

These charges create the nagnetic dipoles.

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Mateo makes a table of the symbols that represent the parts of an electric circuit. What label should Mateo place in the marked

deff fn [24]

The correct option is PROVIDE ELECTRICAL ENERGY.
The marked cell in the given electric circuit is the battery and the function of the battery in electric cells is to provide electrical energy that will drive the electrical current in the electric circuit. A battery in an electric circuit is thus a source of energy for the circuit.

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3 years ago

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Represent a vector of 100 N in North-East direction

andreyandreev [35.5K]

Answer:

please find the attachment to this question.

Explanation:

In this question, we represent the 100N in the North-East direction, but first, we define the vector representation:

It is generally represented through arrows, whose length and direction reflect the magnitude and direction of the arrow points. In this, both size and direction are necessary because the magnitude of a vector would be a number that can be compared to one vector.

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3 years ago

A sphere of mass m" = 2 kg travels with a velocity of magnitude υ") = 8 m/s toward a sphere of mass m- = 3 kg initially at rest,

aleksklad [387]

a) 6.4 m/s

b) 2.1 m

c) $61.6^{\circ}$

d) 14.0 N

e) 4.6 m/s

f) 37.9 N

Explanation:

Since the system is isolated (no external forces on it), the total momentum of the system is conserved, so we can write:

$p_i = p_f\\m_1 u_1 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 2 kg$ is the mass of the 1st sphere

$m_2 = 3kg$ is the mass of the 2nd sphere

$u_1 = 8 m/s$ is the initial velocity of the 1st sphere

$v_1$ is the final velocity of the 1st sphere

$v_2$ is the final velocity of the 2nd sphere

Since the collision is elastic, the total kinetic energy is also conserved:

$E_i=E_k\\\frac{1}{2}m_1 u_1^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Combining the two equations together, we can find the final velocity of the 2nd sphere:

$v_2=\frac{2m_1}{m_1+m_2}u_1=\frac{2(2)}{2+3}(8)=6.4 m/s$

Now we analyze the 2nd sphere from the moment it starts its motion till the moment it reaches the maximum height.

Since its total mechanical energy is conserved, its initial kinetic energy is entirely converted into gravitational potential energy at the highest point.

So we can write:

$KE_i = PE_f$

$\frac{1}{2}mv^2 = mgh$

where

m = 3 kg is the mass of the sphere

v = 6.4 m/s is the initial speed of the sphere

$g=9.8 m/s^2$ is the acceleration due to gravity

h is the maximum height reached

Solving for h, we find

$h=\frac{v^2}{2g}=\frac{(6.4)^2}{2(9.8)}=2.1 m$

Here the 2nd sphere is tied to a rope of length

L = 4 m

We know that the maximum height reached by the sphere in its motion is

h = 2.1 m

Calling $\theta$ the angle that the rope makes with the vertical, we can write

$h = L-Lcos \theta$

Which can be rewritten as

$h=L(1-cos \theta)$

Solving for $\theta$ , we can find the angle between the rope and the vertical:

$cos \theta = 1-\frac{h}{L}=1-\frac{2.1}{4}=0.475\\\theta=cos^{-1}(0.475)=61.6^{\circ}$

The motion of the sphere is part of a circular motion. The forces acting along the centripetal direction are:

- The tension in the rope, T, inward

- The component of the weight along the radial direction, $mg cos \theta$ , outward

Their resultant must be equal to the centripetal force, so we can write:

$T-mg cos \theta = m\frac{v^2}{r}$

where r = L (the radius of the circle is the length of the rope).

However, when the sphere is at the highest point, it is at rest, so

v = 0

Therefore we have

$T-mg cos \theta=0$

So we can find the tension:

$T=mg cos \theta=(3)(9.8)(cos 61.6^{\circ})=14.0 N$

We can solve this part by applying again the law of conservation of energy.

In fact, when the sphere is at a height of h = 1 m, it has both kinetic and potential energy. So we can write:

$KE_i = KE_f + PE_f\\\frac{1}{2}mv^2 = \frac{1}{2}mv'^2 + mgh'$

where:

$KE_i$ is the initial kinetic energy

$KE_f$ is the kinetic energy at 1 m

$PE_f$ is the final potential energy

v = 6.4 m/s is the speed at the bottom

v' is the speed at a height of 1 m

h' = 1 m is the height

m = 3 kg is the mass of the sphere

And solving for v', we find:

$v'=\sqrt{v^2-2gh'}=\sqrt{6.4^2-2(9.8)(1)}=4.6 m/s$

Again, since the sphere is in circular motion, the equation of the forces along the radial direction is

$T-mg cos \theta = m\frac{v^2}{r}$

where

T is the tension in the string

$mg cos \theta$ is the component of the weight in the radial direction

$m\frac{v^2}{r}$ is the centripetal force

In this situation we have

v = 4.6 m/s is the speed of the sphere

$cos \theta$ can be rewritten as (see part c)

$cos \theta = 1-\frac{h'}{L}$

where in this case,

h' = 1 m

L = 4 m

And $r=L=4 m$ is the radius of the circle

Substituting and solving for T, we find:

$T=mg cos \theta + m\frac{v^2}{r}=mg(1-\frac{h'}{L})+m\frac{v^2}{L}=\\=(3)(9.8)(1-\frac{1}{4})+(3)\frac{4.6^2}{4}=37.9 N$

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3 years ago

Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m

ira [324]

Answer:

$\Delta L=15\,mm$

Explanation:

Given:

length of a steel-string, $L=1m$

area of the string, $A=0.5\,mm^2$

Young's modulus of the steel, $Y=2\times 10^{11} Pa$

force of tension on the string, $F=1500\,N$

We have the relation for change in length:

$\Delta L=\frac{F.L}{A.Y}$

$\Delta L=\frac{1500\times 1000}{0.5\times 10^{-6}\times 2\times 10^{11}}$

$\Delta L=0.015m$

$\Delta L=15\,mm$

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3 years ago

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A customer is billed for a service or product, but it gets recorded in the ledger at a lower amount is an example of fraudulent bookkeeping on the income side. This fraud would likely happen in the account receivable department of a company which generates sales for its function<span>. The sale is the primary source of company's operational income.</span>

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3 years ago

basic source of magnetism is a) charged particles alone b)Movement of charged particles c) Magnetic dipoles d)magnetic domains ​

basic source of magnetism is a) charged particles alone b)Movement of charged particles c) Magnetic dipoles d)magnetic domains