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BlackZzzverrR [31]
3 years ago
11

We observe that a small sample of material placed in a non-uniform magnetic field accelerates toward a region of stronger field.

What can we say about the material?
Physics
1 answer:
Aleksandr [31]3 years ago
8 0

Answer:

<em>C) It is either ferromagnetic or paramagnetic</em>

Explanation:

The complete question is given below

We observe that a small sample of material placed in a non-uniform magnetic field accelerates toward a region of stronger field. What can we say about the material?

A) It must be ferromagnetic.

B) It must be paramagnetic.

C) It is either ferromagnetic or paramagnetic.

D) It must be diamagnetic.

A ferromagnetic material will respond towards a magnetic field. They are those materials that are attracted to a magnet. Ferromagnetism is associated with our everyday magnets and is the strongest form of magnetism in nature. Iron and its alloys is very good example of a material that readily demonstrate ferromagnetism.

Paramagnetic materials are weakly attracted to an externally applied magnetic field. They usually accelerate towards an electric field, and form internal induced magnetic field in the direction of the external magnetic field.

The difference is that ferromagnetic materials can retain their magnetization when the externally applied magnetic field is removed, unlike paramagnetic materials that do not retain their magnetization.

In contrast, a diamagnetic material is repelled away from an externally applied magnetic field.

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3 years ago
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a 2000 kg car moving down the road runs into a 5000 kg stationary suv. The car applies a force of 1400 n on the suv what is the
masha68 [24]

Answer:

F_suv= 49050 N

Explanation:

We are told that a 2000 kg car moving down the road runs into a 5000 kg stationary suv. The car applies a force of 1400 N on the suv.

Now, according Newton's first law of motion, an object will continue in it's present state of rest except it is acted upon by an external body.

This means the force acting on the stationary Suv is force of gravity.

Thus; F_suv = 5000 × 9.81

F_suv= 49050 N

3 0
3 years ago
You are on a new planet, and are trying to use a simple pendulum to find the gravity experienced on this new planet. You find th
levacccp [35]

To solve this problem it is necessary to apply the concepts related to the Period based on the length of its rope and gravity, mathematically it can be expressed as

T= 2\pi \sqrt{\frac{L}{g}}

g = Gravity

L = Length

T = Period

Re-arrange to find the gravity we have

g = \frac{4\pi^2 L}{T^2}

Our values are given as

L = 0.35m\\T = 2s\\

Replacing we have

g = \frac{4\pi^2 L}{T^2}

g = \frac{4\pi^2 0.35}{2^2}

g = 3.45 m/s^2

Therefore the correct answer is C.

4 0
3 years ago
Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.250 A/s,
sdas [7]

Explanation:

Given that,

Induced emf \epsilon= 1.65\times10^{-3}\ V

Rate of current = 0.250 A/s

Number of turns =30

(a). We need to calculate the mutual inductance of the pair of coils

Using formula of the mutual inductance

M=\dfrac{\epsilon}{|\dfrac{\Delta i}{\Delta t}|}

M=\dfrac{1.65\times10^{-3}}{0.250}

M=0.0066=6.6\times10^{-3}\ Hz

M=6.6\ mHz

The mutual inductance of the pair of coils is 6.6 mHz.

(b). We need to calculate the flux through each turn

Using formula of flux

\phi_{B}=\dfrac{Mi}{N}

Put the value into the formula

\phi_{B}=\dfrac{6.6\times10^{-3}\times1.25}{30}

\phi_{B}=0.000275 =2.75\times10^{-4}\ Wb

The flux through each turn is 2.75\times10^{-4}\ Wb

(c). We need to calculate the magnitude of the induced emf in the first coil

Using formula of induced emf

\epsilon=M|\dfrac{\Delta i_{2}}{\Delta t}|

\epsilon=6.6\times10^{-3}\times0.3

\epsilon=0.00198 =1.98\times10^{-3}\ mV

The magnitude of the induced emf in the first coil is 1.98\times10^{-3}\ mV

Hence, This is the required solution.

8 0
3 years ago
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A light ray just grazes the surface of the Earth (M = 6.0 × 10 24 kg, R = 6.4×10 6 m). Through what angle α is the light ray ben
grandymaker [24]

Answer:

(a). The deflection angle is 2.77\times10^{-9}\ rad

(b). The deflection angle is  3.95\times10^{-4}\ rad

(c). The deflection angle is 7.41\times10^{-1}\ rad

Explanation:

Given that,

Mass of earth M_{e}=6.0\times10^{24}\ kg

Radius of earth R_{e}=6.4\times10^{6}\ m

Mass of white dwarf M=2.0\times10^{30}\ kg

Radius of white dwarf R=1.5\times10^{7}\ m

Mass of Neutron M=3.0\times10^{30}\ kg

Radius of neutron R=1.2\times10^{4}\ m

We need to calculate the deflection angle for earth

Using formula of angle

\alpha=\dfrac{4G M}{c^2R}

Where, R = radius

G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

\alpha=\dfrac{4\times6.67\times10^{-11}\times6.0\times10^{24}}{(3\times10^{8})^2\times6.4\times10^{6}}

\alpha=2.77\times10^{-9}\ rad

The deflection angle is 2.77\times10^{-9}\ rad

We need to calculate the deflection angle for white dwarf

Using formula of angle

\alpha=\dfrac{4G M}{c^2R}

Put the value into the formula

\alpha=\dfrac{4\times6.67\times10^{-11}\times2.0\times10^{30}}{(3\times10^{8})^2\times1.5\times10^{7}}

\alpha=3.95\times10^{-4}\ rad

The deflection angle is 3.95\times10^{-4}\ rad

We need to calculate the deflection angle for neutron star

Using formula of angle

\alpha=\dfrac{4G M}{c^2R}

Put the value into the formula

\alpha=\dfrac{4\times6.67\times10^{-11}\times3.0\times10^{30}}{(3\times10^{8})^2\times1.2\times10^{4}}

\alpha=7.41\times10^{-1}\ rad

The deflection angle is 7.41\times10^{-1}\ rad

Hence, This is the required solution.

3 0
4 years ago
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