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2 years ago

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The properties of some rare earth elements make them essential ingredients in electronics. Rare earth elements appear together i

n the same general area of the periodic table of the elements. The graph shows two properties of several rare earth elements.
Is it accurate to say that the rare earth elements, as a group, share the same properties? Support your answer using the information in the graph.

2 answers:

Vera_Pavlovna [14]2 years ago

7 0

Answer:

I'm pretty sure it is accurate.

Explanation:

USPshnik [31]2 years ago

4 0

Answer:

It is Inaccurate.

Explanation:

The graph is inaccurate because the rare-earth elements shown on the graph all have varying melting and boiling points, some more extreme than others. seeing how melting/boiling point is the only property compared between the rare earth elements, it is inaccurate to say that these elements share the same properties.

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A 8.8 cm diameter circular loop of wire is in a 1.04 T magnetic field. The loop is removed from the field in 0.30 s . Assume tha

denis23 [38]

Answer:

0.021 V

Explanation:

The average induced emf (E) can be calculated usgin the Faraday's Law:

$E = - \frac{N*\Delta \phi}{\Delta t}$

<u>Where:</u>

<em>N = is the number of turns = 1 </em>

<em>ΔΦ = ΔB*A </em>

<em>Δt = is the time = 0.3 s </em>

<em>A = is the loop of wire area = πr² = πd²/4 </em>

<em>ΔB: is the magnetic field = (0 - 1.04) T </em>

Hence the average induced emf is:

$E = - \frac{\Delta B*A}{\Delta t} = - \frac{(0- 1.04 T) \pi (0.088 m)^{2}}{4*0.3 s} = 0.021 V$

Therefore, the average induced emf is 0.021 V.

I hope it helps you!

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3 years ago

A 1,500-kg car accelerates along a flat track to a speed of 20 m/s. how much does it's gravitational potential energy increases

lesya [120]

Look at bitesizes physics section, they have all the information you need to complete this question.

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2 years ago

Determine the magnitude of the effective value of g⃗ at a latitude of 60 ∘ on the earth. assume the earth is a rotating sphere.

dezoksy [38]

In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr)
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2 years ago

A flat coil having 160 turns, each with an area of 0.20 m 2, is placed with the plane of its area perpendicular to a magnetic fi

S_A_V [24]

Answer:

10.2 Watt

Explanation:

$N$ = number of turns in flat coil = 160

$A$ = area = 0.20 m²

B₀= initial magnetic field = 0.40 T

$B$ = final magnetic field = - 0.40 T

Change in magnetic field is given as

ΔB = B - B₀ = - 0.40 - 0.40 = - 0.80 T

$t$ = time taken for the magnetic field to change = 2.0 s

Induced emf is given as

$E = \frac{- N A \Delta B}{t}$

$E = \frac{- (160) (0.20) (- 0.80)}{2}$

$E$ = 12.8 volts

$R$ = Resistance of the coil = 16 Ω

Power is given as

$P = \frac{E^{2}}{R}$

$P = \frac{(12.8)^{2}}{16}$

$P$ = 10.2 Watt

6 0

2 years ago

9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o

Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

$m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.$

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

$F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).$

Now the force of friction is $F_s=\mu*N,$ where $N$ is the normal force and from the diagram it is $F_y=mg*cos(\theta).$

Thus $F_s=\mu*N=\mu*mg*cos(\theta).$

Therefore,

$F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).$

Substituting the value for $F_H,m,\mu, and \:\theta$ we get:

$F_{tot}= -38.63N.$

Now acceleration is simply

$a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.$

The negative sign indicates that the acceleration is directed up the incline.

Part B:

$d=\frac{1}{2} at^2$

Which can be rearranged to solve for t:

$t=\sqrt{\frac{2d}{a} }$

Substitute the value of $d=0.50m$ and $a=7.7m/s$ and we get:

$t=0.36s.$

which is our answer.

Notice that in using the formula to calculate time we used the positive value of $a$ , because for this formula absolute value is needed.

5 0

3 years ago