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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 small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of
ehidna [41]

Answer:

Explanation:

Given

Speed of block at bottom is v=3.8\ m/s

Distance traveled s=6\ m

initial velocity is zero

using equation of motion

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

(3.8)^2-0=2\times a\times 6

a=1.203\ m/s^2

when it is 3 m from top then

v^2-u^2=2as

v^2-0=2\times 1.203\times 3

v=2.68\ m/s      

5 0
3 years ago
Can somebody help me understand this
Komok [63]
I think it’s C. Marshmallow and toothpicks are used to show the composition of a water molecule
6 0
3 years ago
A spring has a natural length of 0.5 m and was stretched by 0.02 m. if the spring had a resultant energy of 0.5 j what is the sp
Anna71 [15]

\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}

Explanation:

       Natural length of a spring is 0.5\text{ }m. The spring is streched by 0.02\text{ }m. The resultant energy of the spring is 0.5\text{ }J.

       The potential energy of an ideal spring with spring constant k and elongation x is given by \dfrac{1}{2}kx^{2}.

       So, in the current problem, the natural length of the spring is not required to find the spring constant k.

       \text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}

∴ The spring constant of the spring = 2500\text{ }\frac{kg}{s^{2}}

4 0
3 years ago
(a) A long, straight solenoid has N turns, uniform cross-sectional area A, and length l. Show that the inductance of this soleno
Paul [167]

Answer:

a. L = μ₀AN²/l b. 1.11 × 10⁻⁷ H

Explanation:

a. The magnetic flux through the solenoid, Ф = NAB where N = number of turns of solenoid, A = cross-sectional area of solenoid and B = magnetic field at center of solenoid = μ₀ni where μ₀ = permeability of free space, n = number of turns per unit length = N/l where l = length of solenoid and i = current in solenoid.

Also, Li = Ф where L = inductance of solenoid.

So, Li = NAB

= NA(μ₀ni)

= NA(μ₀Ni/l)

Li = μ₀AN²i/l

dividing both sides by i, we have

So, L = μ₀AN²/l

b. The self- inductance, L = μ₀AN²/l where

A = πd²/4 where d = diameter of solenoid = 0.150 cm = 1.5 × 10⁻³ m, N = 50 turns, μ₀ = 4π × 10⁻⁷ H/m and l = 5.00 cm = 5 × 10⁻² m

So, L = μ₀AN²/l

L = μ₀πd²N²/4l

L = 4π × 10⁻⁷ H/m × π(1.5 × 10⁻³ m)²(50)²/(4 × 5 × 10⁻² m)

L = 11,103.3 × 10⁻¹¹ H

L = 1.11033 × 10⁻⁷ H

L ≅ 1.11 × 10⁻⁷ H

6 0
3 years ago
A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the
Novay_Z [31]

Answer:

It will be cut in half

Explanation:

The diffraction of a slit is given by the formula

a sin θ = m where

a = width of the slit,

λ = wavelength and

m = integer that determines the order of diffraction.

Next we divide both sides by a, we have

sin θ = m λ / a

Also, recall that

a’ = 2 a

Then we substitute in the previous equation

2asin θ' = m λ, if divide by 2a, we have

sin θ' = (m λ / 2a).

Now again, from the first equation, we said that sin θ = m λ / a, so we substitute

sin θ ’= sin θ / 2

Then we use trigonometry to find the width, we say

tan θ = y / L

Since the angle is small, we then have

tan θ = sin θ / cos θ

tan θ = sin θ, this then means that

sin θ = y / L

we will then substitute

y’ / L = y/L 1/2

y' = y / 2

this means that when the slit width is doubled the pattern width will then be halved

4 0
3 years ago
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