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3 years ago

There is a simple pattern for determining if a binary number is odd. what is it and why does this pattern occur

2 answers:

6 0

There is a simple pattern for determining if a binary number is odd. This is when the number ends in 1. This pattern occured because they are defined by an odd word.

<h3>Further explanation </h3>

The binary number is a number that expressed in the base-2 numeral system or binary numeral system, that uses only two symbols, typically "0" and "1".

If the number ends in 1, it will be odd, whereas if the number ends in 0, it's even. Because the first digit can only be 0 or 1, the rest of the digits are multiplied by two. To convert a binary number to base 2k, split it into groups of k digits by adding leading 0s if necessary, then convert each group to base 2k.

Example:

1001 represents an odd number (9)
1000 represents an even number (8)
1101 represents an odd number (13)
10010 represents an even number (18).

This pattern odd occured because they are defined by an odd word.They are numbers, that are by definition is too many things that they become odd, and also because they look odd again even when split into two

<h3>Learn more</h3>

Learn more about binary number brainly.com/question/8604586
Learn more about pattern brainly.com/question/554926
Learn more about odd binary number brainly.com/question/5625911

<h3>Answer details</h3>

Grade: 9

Subject: physics

Chapter: Computer Science

Keywords: binary number, pattern, odd binary number, even binary number, Computer Science

frez [133]3 years ago

5 0

The answer to this question would be the least number (right most number).

If you add an even number with even number, the result should be even. But if there is just one odd number added, it will be odd.
In binary, all the number is multiplied by $2^{x}$ . It is pretty clear that all the number should be even beside the first/lowest multiplier $2^{0}$ which has potential to be 1.

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I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

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

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

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5 0

3 years ago

The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =

luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time = $\frac{\textup{Distance}}{\textup{Velocity}}$

Time = $\frac{\textup{0.016}}{\textup{23}}$

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

a = $\frac{q\times8.20\times10^4}{1\times10^{-11}}$

Now from the Newton's equation of motion

$d=ut+\frac{1}{2}at^2$

where,

d is the distance

u is the initial speed

a is the acceleration

t is the time

$0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2$

q = 9.98 × 10⁻⁹ C

4 0

3 years ago

An object is placed on a surface. A student tries to apply various combinations of forces on the object. Which pair of forces wi

AVprozaik [17]

Answer:

See Explanation

Explanation:

The question is incomplete, as there are no diagrams or options to provide more information to the question.

The general explanation is as follows:

For the object not to move

(1): The forces acting on the object must opposite each other. i.e. if force A acts at the right (or positive direction), force B will act at the left (or negative direction).

(2) The two forces must be equal.

So, for instance:

If the pair of forces are 5N and 5N in opposite directions, the object wil not move.

However, if one of the forces is greater, the object will move towards the direction of the greater force.

7 0

3 years ago

A swing has a period of 10 seconds. What is its frequency ?

kow [346]

Frequency = 1 / (period)

Frequency = 1 / (10 seconds) = (1/10) ( / second) = 0.1 per second = <em>0.1 Hz</em>.

4 0

3 years ago

Over 4 seconds, a car's momentum decreases by 1000 kg m/s how much force did it take to make this happen?

kotykmax [81]

Answer:

250N

Explanation:

Given parameters:

Time = 4s

Momentum = 1000kgm/s

Unknown:

Force = ?

Solution:

To solve this problem, we use Newton's second law of motion;

Ft = Momentum

F is the force

t is the time

So;

F x 4 = 1000kgm/s

F = 250N

6 0

3 years ago