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Artyom0805 [142]
3 years ago
8

The densities of the elements K, Ca, Sc, and Ti are 0.86, 1.5, 3.2, and 4.5 g/cm³, respectively. One of these elements crystalli

zes in a body-centered cubic structure; the other three crystallize in a face-centered cubic structure.
Which one crysta llizes in the body-centered cubic structure? Justify your answer
Physics
1 answer:
IgorC [24]3 years ago
4 0

Answer:

K

Explanation:

Given that

Elements                                    Density

K                                                  0.86 g/cm³

Ca                                                 1.5 g/cm³

Sc                                                  3.2 g/cm³

Ti                                                    4.5 g/cm³

As we know that density of the BCC ( body centered cubic )element is lower than the FCC(Face centered cubic )element.

In the given problem only K have lower density all of these element that is why is in BCC and all others are in FCC.

The packing efficiency of the FCC structure element is higher than the BCC element that is why FCC is more denser than BCC.And also FCC having good packing of element.

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A metal have a nice day
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3 years ago
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A coil 3.55 cm in radius, containing 470 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.2
slavikrds [6]

The Electric current is 1.11* 10^{-4}A


Given that the coil's radius is 3.55 cm (0.35 m),

The formula for the coil's area is A = r2 A = (3.14) (0.35)2 = 0.005024 m2.

R = Resistance = 600 N = Number of spins = 500 B = Magnetic field = (0.0120)

t + (3 x 10⁻⁵) t⁴

The number t = 5 is substituted for taking the derivative at both the induced current and the electric current.

The Electric current is therefore 1.11* 10^{-4}A
Electric current - The rate of electron passage in a conductor is known as electric current. The ampere is the electric current's SI unit. Electrons are little particles that are part of a substance's molecular structure. These electrons can be held loosely or securely depending on the situation.

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7 0
2 years ago
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A motorist drives north for 35.0 minutes at 85.0 kph. He then stops for 15.0 minutes. The motorist then drives 130.0 km in 2.0 h
o-na [289]

Answer:

A. 216.36 \frac{km}{h}

B. 96.56 \frac{km}{h}

Explanation:

Let s_{1} be the distance in first part.

s_{1} = velocity × time

s_{1} = 85 × \frac{35}{60}

s_{1} = 49.58 km

Let s_{2} be the distance in first part.

s_{2} = 130 km

Average velocity = \frac{Total displacement}{Total time}

When second leg of the trip is

A. Toward north

Average velocity = \frac{[tex] s_{1} + s_{2}}{Total time}  [/tex]

Average velocity = \frac{130+49.58}{0.25+0.58}

Average velocity =216.36 \frac{km}{h}

B. Toward south

Average velocity = \frac{[tex] s_{1} - s_{2}}{Total time}  [/tex]

Average velocity = \frac{130-49.58}{0.25+0.58}

Average velocity =96.56 \frac{km}{h}


3 0
3 years ago
Jack and Jill are maneuvering a 3100 kg boat near a dock. Initially the boat's position is < 2, 0, 3 > m and its speed is
liraira [26]

Answer:

(a) work done by Jack is -2200J

(b) work done by Jill is 0

Explanation:

Given;

Initially boat's position =  < 2, 0, 3 > m

Final boat's position = < 6, 0, 1 > m

change in position = < 4, 0, -2 >

Force exerted by Jack =  < -440, 0, 220 > N

Force exerted by Jill = < 170, 0, 340 > N

Part (a) How much work does Jack do?

work done = force x distance

Jack's work = < -440, 0, 220 > N x < 4, 0, -2 > m

                    = [(-440*4) + (0*0) + (220*-2)]

                    = -1760 J - 440 J

                    = -2200J

Part (b) How much work does Jill do?

Jill's work = < 170, 0, 340 > N x < 4, 0, -2> m

                 = [(170*4) + (0*0) + (340*-2)]

                  = 680 J - 680 J

                   = 0

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4 years ago
in a football game, the kicker kicks a football a horizontal distance of 43 yards if the ball lands 3.9 seconds later, what is t
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Answer:

10s

Explanation:

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