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Nuetrik [128]
3 years ago
11

A knight moves on a chessboard two squares up, down, left, or right followedby one square in one of the two directions perpendic

ular to the first part of the move (i.e.,the move is L-shaped). Suppose the knight is on an unbounded board at square (0,0) and wewish to move it to square (x,y) in the smallest number of moves. (For example, to move from(0,0) to (1,1) requires two moves. The knight can move to board locations with negativecoordinates.)

Physics
1 answer:
Contact [7]3 years ago
5 0

Answer:

Explanation:

Check attachment for solution.

Generally the movement of the knight is L, i.e (2,1), (1,2),(-1,2),(1,-2) etc.

So using Pythagoras theorem

x^2+y^2=1^2+2^2

x^2+y^2=5

Then, the knight will make one movement when the displacement is √5.

So let take a look at other positions

(1,0),(0,1),(-1,0), (0,-1).

Then, for us to have this kind of movement, the knight has to make 3 movements.

When the displacement is 1, then it will make 3 movement.

Let examine other positions

(2,2) or(-2,-2) or (2,-2) or (-2,2)

When the displacement is √8

Then, the movement of the knight is 4.

Let examine other points

(2,0) or (0,2) or (-2,0) or (0,-2)

When the displacement is 2.

The knight make 2 movement

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The dimensions of a cylinder are changing, but the height is always equal to the diameter of the base of the cylinder. If the he
VashaNatasha [74]

Answer:

dV/dt  = 9 cubic inches per second

Explanation:

Let the height of the cylinder is h

Diameter of cylinder = height of the cylinder = h

Radius of cylinder, r = h/2

dh/dt = 3 inches /s

Volume of cylinder is given by

V = \pi r^{2}h

put r = h/2 so,

V = \pi \frac{h^{3}}{4}

Differentiate both sides with respect to t.

\frac{dV}{dt}=\frac{3h^{2}}{4}\times \frac{dh}{dt}

Substitute the values, h = 2 inches, dh/dt = 3 inches / s

\frac{dV}{dt}=\frac{3\times 2\times 2}{4}\times 3

dV/dt  = 9 cubic inches per second

Thus, the volume of cylinder increases by the rate of 9 cubic inches per second.

6 0
3 years ago
How is light refracted inside a crystal ball ?
Bingel [31]
The light refects ofthe sides 
7 0
3 years ago
Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te
seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy  5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant \gamma =1.4

specific heat is given as =\frac{\gamma R}{\gamma -1}

gas constant =287 J⋅kg−1⋅K−1

Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}

specific heat at constant volume

Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}

change in internal energy = Cv(T_2 -T_1)

                            \Delta U = 717.5 (800-295)  = 3.62*10^5 J kg^{-1}

change in enthalapy \Delta H = Cp(T_2 -T_1)

                                 \Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}

change in entropy

\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})

\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})

\Delta S = 382.79 J kg^{-1} K^{-1}

7 0
3 years ago
A ball thrown horizontally at vi = 30.0 m/s travels a horizontal distance of d = 55.0 m before hitting the ground. from what hei
tekilochka [14]
Assume no air resistance, and g = 9.8 m/s².

Let
x =  angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity

The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x)  s

With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
 55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0

Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°

The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s

If t = 5.8096 s,
  u*t = 9.467*5.8096 = 55 m (Correct)
or
 u*t = 28.469*15.8096 = 165.4 m (Incorrect)

Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s

The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m

Answer: h = 110.4 m

7 0
3 years ago
What is a literature review?<br>​
almond37 [142]

Answer: A literature review consists of an overview, a summary, and an evaluation (“critique”) of the current state of knowledge about a specific area of research.

Explanation:

7 0
2 years ago
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