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lorasvet [3.4K]
3 years ago
5

Diamagnetic materialsA) have small negative values of magnetic susceptibility.B) are those in which the magnetic moments of all

electrons in each atom cancel.C) experience a small induced magnetic moment when placed in an external magnetic field.D) exhibit the property of diamagnetism independently of temperature.E)are described by all
Physics
1 answer:
White raven [17]3 years ago
4 0

Answer:

C) experience a small induced magnetic moment when placed in an external magnetic field.

Explanation:

Diamagnetics materials are those that experience a small induced magnetic moment when placed in an external magnetic field. These materials, such as bismuth, copper, silver and lead, have elementary magnets in their compositions. When they are exposed to an external magnetic cap, these elemental magnets tend to follow an orientation contrary to the external magnetic field. As a result, a magnetic field is created in the opposite direction to the external magnetic field.

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Bats are capable of navigating using the earth's field-a plus for an animal that may fly great distances from its roost at night
Nataliya [291]

Answer:

The correct magnitude of the coil's magnetic field= =50μT

Explanation :

The magnetic field takes place as a result of movement of charge I e current, can also occurs from magnetised material, magnetic field as a result of charge movement can be deducted using right hand grip rule.

At the equator magnetic field lines are parallel towards the earth's surface and the angle of inclination of the magnetic lines of force at the horizontal position is referred to as the angle of dip at the point.

As the current is produced then the varying magnetic field is opposed ,then there is induced current when the coil is positioned at varying magnetic field.

Given from the question,

angle below horizontal θ=60-degree

The Earth's magnetic field B=50μT

The horizontal magnetic field can be expressed in terms of the formula below;

BH=Bcos⁡θ

B(H) = earth's horizontal component of magnetic field

Ø is the angle between coil's field

B=the magnetic field in Tesla

Then,

BH=50μT×cos60∘

=50μT× 0.5

As the current is passed through the coil to produce a field , when combined with the earth's field, which creates a net field with the same strength and dip angle (60 degrees below horizontal) as the earth's field.

It can be deducted that B has the same magnitude and angle which makes the those vertical component to cancel each other since they are the same.

For the magnetic field to be pointed out at North direction, we can calculate the corrected magnetic field using the formula below

Bc=2BH

Bc=2×50μT× 0.5=50μT

The correct magnitude of the coil's magnetic field= =50μT

4 0
4 years ago
Which of the following is equal to the area under a velocity-time graph
Ksenya-84 [330]

-- The area under a velocity/time graph, between two points in time, is the difference in displacement during that period of time.

-- The area under a speed/time graph, between two points in time, is the distance covered during that period of time.

5 0
3 years ago
Read 2 more answers
How does the force on the rifle compare with the force on the bullet, and why?
I am Lyosha [343]
They are equal and opposite (action and reaction)
6 0
4 years ago
A freight car of mass M contains a mass of sand m. At t = 0 a constant horizontal force F is applied in the direction of rolling
Veronika [31]

Answer:

Amount of linear movement

Explanation:

Our system is defined by the rate of change in mass that

leaves the car \Delta m_ {s} , this happens during a time interval

[t, t + \Delta t], in addition to freight car and sand at time t.

In this way we need to define the two states:

State 1,

consider t, m_ {c} (t) + \Delta m_ {s} and V.

State 2,

consider t + \Delta t, m_ {c} (t), V + V \Delta V

In this state is the mass of sand output, which

is composed of

\Delta m_{s}, V + \Delta V

In this way we define the Linear movement in x, like this:

p_ {x} (t) = (\Delta m_ {s} + m_ {c} (t)) v

p_ {x} (t+\Delta t) = (\Delta m_ {s} + m_ {c} (t)) (v + \Delta v)

m_ {c} (t) = m_ {c, 0} - bt = m_ {c} + m_ {s} -bt

In this way we proceed to obtain the Force

F =\lim_{\Delta t \rightarrow 0} \frac {p_x (t + \Delta t) -p_ {x} (t)} {\Delta t}

F = lim_{\Delta t \rightarrow 0} m_ {c} (t) \frac {\Delta v} {\Delta t} + lim_{\Delta t \rightarrow 0} m_ {s} (t) \frac {\Delta v} {\Delta T}

Since the mass of the second term becomes 0, the same term is eliminated, thus,

F = m_ {c} (t) \frac {dv} {dt}

\int\limit ^ {v (t)} _ {v = 0} dv = \int\limit^t_0 \frac{Fdt} {m_ {c} + m_ {s} -bt}

V (t) = - \frac {F} {b} ln (\frac {m_c + m_s-bt} {m_c + m_s})

3 0
3 years ago
A 0.290 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 30.0 pC charge on it
mojhsa [17]

Answer:

Therefore,

The potential (in V) near its surface is 186.13 Volt.

Explanation:

Given:

Diameter of sphere,

d= 0.29 cm

radius=\dfrac{d}{2}=\dfrac{0.29}{2}=0.145\ cm

r = 0.145\ cm = 0.145\times 10^{-2}\ m

Charge ,

Q = 30.0\ pC=30\times 10^{-12}

To Find:

Electric potential , V = ?

Solution:

Electric Potential at point surface is given as,

V=\dfrac{1}{4\pi\epsilon_{0}}\times \dfrac{Q}{r}

Where,  

V= Electric potential,  

ε0 = permeability free space = 8.85 × 10–12 F/m

Q = Charge  

r = Radius  

Substituting the values we get

V=\dfrac{1}{4\times 3.14\times 8.85\times 10^{-12}}\times \dfrac{30\times 10^{-12}}{0.145\times 10^{-2}}

V=\dfrac{30}{16.117\times 10^{-2}}=186.13\ Volt

Therefore,

The potential (in V) near its surface is 186.13 Volt.

3 0
4 years ago
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