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CaHeK987 [17]
3 years ago
12

Define Miller index, and the recipe for how to find the Miller index for a particular lattice plane

Physics
1 answer:
Iteru [2.4K]3 years ago
4 0

Answer:

 Miller index:

      This is the set of number which is used to define the position of face any crystal.We know that crystal plane is the plane where large number of molecule are present inside this plane.

Lets take r is the vector which is passing through the origin of lattice

r = a i + b j + c k

So the intercept on the plane x is a unit ,on the y plane is b unit and on the z plane is c unit.

Now when we take the reciprocal of these intercept we get 1/a, 1/b  , and 1/c.These reciprocal of intercept is know as Miller index.

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You drive a car east on the highway at 26 m/s. Another car passes you moving east traveling at 32 m/s. How fast do you view the
tamaranim1 [39]

The speed of the car passing you is 6 m/s while car is moving 6 m/s behind the car.

<h3>Relative velocity of the car</h3>

The speed of the car passing you is determined by applying relative velocity principle as shown below;

Vr = Va - Vb

Vr = 26 m/s - 32 m/s

Vr = -6 m/s

Thus, the speed of the car passing you is 6 m/s while car is moving 6 m/s behind the car.

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3 0
2 years ago
A plane opens a parachute after it lands inorder to
Elis [28]

Answer:

When the parachute opens, the air resistance increases. The skydiver slows down until a new, lower terminal velocity is reached.

8 0
2 years ago
Read 2 more answers
A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t
notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is 1.37 m/s^2

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, F_f, backward

The frictional force can be written as

F_f = \mu mg

where

\mu=0.03 is the coefficient of kinetic friction

m=150 kg is the mass of the motorbike

g=9.8 m/s^2 is the acceleration of gravity

Therefore the net force is given by

\sum F = F - F_f = F - \mu mg

And substituting, we find

\sum F=250 - (0.03)(150)(9.8)=205.9 N

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

\sum F = ma

where

m is the mass

a is the acceleration

In this problem, we have

\sum F = 205.9 N is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2

C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

a=1.37 m/s^2

s = 350 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s

4)

For this part of the problem, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the elapsed time

Here we have:

v = 5.2 m/s

u = 0

a=1.37 m/s^2

Solving for t, we find

t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s

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6 0
3 years ago
PART ONE
Lina20 [59]

Explanation:

Make a table, listing the x and y coordinates of each square's center of gravity and its mass.  Multiply the coordinates by the mass, add the results for each x and y, then divide by the total mass.

\left\begin{array}{ccccc}x&y&m&xm&ym\\\frac{a}{2} &\frac{a}{2} &10&5a&5a\\\frac{3a}{2}&\frac{a}{2}&70&105a&35a\\\frac{a}{2}&\frac{3a}{2}&80&40a&120a\\\frac{3a}{2}&\frac{3a}{2}&50&75a&75a\\&\sum&210&225a&235a\\&&Avg&\frac{15a}{14}&\frac{47a}{42}\end{array}\right

The x-coordinate of the center of gravity is 15/14 a.

The y-coordinate of the center of gravity is 47/42 a.

4 0
3 years ago
A fisherman is fishing from a bridge and is using a "44.0-N test line." In other words, the line will sustain a maximum force of
avanturin [10]

Answer:

(a) W= 44N

(b)W= 31.65 N

Explanation:

Data

T=44 N : Maximum force that the rope can withstand without breaking

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

(a)  We apply the formula (1) at constant speed , then, a=0

W: heaviest fish that can be pulled up vertically

∑F = 0

T-W =0  

W = T

W= 44N

(b)  We apply the formula (1) , a= 1.26 m/s²

W: heaviest fish that can be pulled up vertically

W= m*g

m= W/g

g= 9.8 m/s² : acceleration due to gravity

∑F = 0

T-W = m*a

T= W+(W/g)*a

44=W*(1+1/9.8)* (1.26 )

44= W* 1.39

W= 44/1.39

W= 31.65 N

7 0
3 years ago
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