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ohaa [14]
3 years ago
11

(a) Compute the radius r of an impurity atom that will just fit into an FCC octahedral site in terms of the atomic radius R of t

he host atom (without introducing lattice strains). Entry field with correct answer 0.414 R
Physics
1 answer:
ratelena [41]3 years ago
5 0

Answer:

radius r is  0.414 R

Explanation:

Given data

FCC octahedral site

atomic radius R

to find out

radius r

solution

we know that at FCC octahedral

length of side = 2 R + 2r

and by pythagorean theorem

a = 2√2R

here a = 2R + 2r

so 2R + 2r = 2√2R

so r = ( √2R )- R

r = 0.414 R

so  radius r is  0.414 R

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At divergent boundaries, hot mantle rock rises and _____ occurs.
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When rock rises<span>, they decrease in pressure causes </span>hot mantle rock<span> to melt and form magma. In plate tectonics, </span>divergent boundaries occur<span> when plates pull apart.</span>
5 0
2 years ago
Read 2 more answers
To pull an old stump out of the ground, you and a friend tie two ropes to the stump. You pull on it with a force of 500 N to the
zhannawk [14.2K]

Answer:

C. less than 950 N.

Explanation:

Given that

Force in north direction F₁ = 500 N

Force in the northwest F₂ = 450 N

Lets take resultant force R

The angle between force = θ

θ = 45°

The resultant force R

R=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta}

R=\sqrt{500^2+450^2+2\times 450\times 500\times cos\theta}

R= 877.89 N

Therefore resultant force is less than 950 N.

C. less than 950 N

Note- When these two force will act in the same direction then the resultant force will be 950 N.

8 0
3 years ago
An object is thrown upward from the top of a 144​-foot building with an initial velocity of 128 feet per second. The height h of
ankoles [38]

Answer:

After 9 seconds the object reaches ground.

Step-by-step explanation:

We equation of motion given as h = -16t²+128t+144,

We need to find in how many seconds will the object hit the ground,

That is we need to find time when h = 0

                        0 = -16t²+128t+144

                       16t²-128t-144= 0

                       t=\frac{-(-128)\pm \sqrt{(-128)^2-4\times 16\times (-144)}}{2\times 16}\\\\t=\frac{128\pm \sqrt{25600}}{32}\\\\t=\frac{128\pm 160}{32}\\\\t=9s\texttt{ or }t=-1s

Negative time is not possible, hence after 9 seconds the object reaches ground.

8 0
2 years ago
An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.
Georgia [21]

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

The \ weight, W  =G\dfrac{M \times m}{R^{2}} = 800 \ N

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

W_1   =G\dfrac{\dfrac{M}{2}  \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2}  \times m \times 4}{ R ^{2}} = 2 \times G \times  \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

7 0
2 years ago
If you push on a wall with a force of 200 N, with what force does the wall push back?
PSYCHO15rus [73]
200 N, that is if the force is balanced and the wall doesn't move
8 0
2 years ago
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