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3 years ago

Which of the following characteristics are displayed by hard magnetic materials?

Physics

1 answer:

irakobra [83]3 years ago

4 0

Answer:

Characteristics are displayed by hard magnetic materials:

B. relatively large hysteresis loop.

D. Magnetization and demagnetization require relatively high applied fields.

<u>Explanation</u>:

They are otherwise called as the permanent magnets, they have the ability to retain is quality even after being magnetised. They have the ability of greater hysteresis loop, it is the relation between the induced magnetic flux to that of magnetizing force. When it is high gives the hard magnetic material. They have high retentivity and coercivity. This means the driving field during magnetization is null. To get coercivity the driving field should be made in the opposite direction.

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Assuming the accleration applied was constant, we have

$v=v_0+at\implies v_0+20.0\,\dfrac{\mathrm m}{\mathrm s}=v_0+a(0.10\,\mathrm s)$

$\implies20.0\,\dfrac{\mathrm m}{\mathrm s}=a(0.10\,\mathrm s)$

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Answer:

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3 years ago

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following case

Murljashka [212]

Answer:

The answer is below

Explanation:

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the

cable for the following cases:

a. The load moves downward at a constant velocity

b. The load accelerates downward at a rate 0.4 m/s??

C. The load accelerates upward at a rate 0.4 m/s??

Solution:

Acceleration due to gravity (g) = 10 m/s²

a) Given that the mass of the crane (m) is 140 kg. If the load moves downward, the tension (T) is given by:

mg - T = ma

Since the load has a constant velocity, hence acceleration (a) = 0. Therefore:

mg - T = m(0)

mg - T = 0

T = mg

T = 140(10) = 1400 N

T = 1400 N

b) If the load moves downward, the tension (T) is given by:

mg - T = ma

T = mg - ma = m(g - a)

T = 140(10 - 0.4) = 140(9.96) = 134.4

T = 134.4 N

c) If the load moves upward, the tension (T) is given by:

T - mg = ma

T = ma + mg = m(a + g)

T = 140(0.4 + 10) = 140(10.4)

T = 145.6 N

2) To find the distance (s) if the load move from rest (u= 0) and accelerates for 20 seconds (t = 20). We use:

s = ut + (1/2)gt²

s = 0(20) + (1/2)(10)(20)²

s = 2000 m

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3 years ago

Calculate the speed of a proton after it accelerates from rest through a potential difference of 380 v .

nikitadnepr [17]

The magnitude of the change in potential energy is equal to the kinetic energy,

$\left | \Delta U \right |= K$

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

Here, V is potential difference, q is charge, m is mass and v is velocity.

We can also write,

$v=\sqrt{\frac{2qV}{m} }$

Given $V=380 V$

Substituting this value with mass of proton, $m=1.672\times10^{-27} kg$ and charge of proton, $q= 1.6\times10^{-19} C$ we get

$v=\sqrt{\frac{2\times1.6\times10^{-19}C \times 380V}{1.672\times10^{-27} kg} }\\\\v= 727.27\times10^{8} m/s =7.27\times10^{6}m/s$

Therefore, the speed of proton is $7.27\times10^{6}m/s$ .

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