Information that is easily converted into numbers and is stored as a on and off signals is _____ information.

Answers

Answer 1

Answer:

"Binary" information

Related Questions

Convert 2.4 milimetre into metre
In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force of 48 N to the input piston, which has a radius r1. As a result, the output plunger, which has a radius r2, applies a force to the car. The ratio r2/r1 has a value of 9.0. Ignore the height difference between the input piston and output plunger and determine the force that the output plunger applies to the car.
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A pair of thin spherical shells with radius r and R, r < R are arranged to share a center. What is the capacitance of the system. If a potential difference V is created between the shells, how much energy is stored between them?
For the system shown below, what is the critical angle (angle at which the system just begins to move)? Assume that the coefficient of friction between all flat surfaces is 0.0500 and that the pulley is frictionless. The mass of m1 is 76.00 kg and the mass of m2 is 194.00 kg. Express your answer in radians.

The electrons in the beam of a television tube have an energy of 19.0 keV. The tube is oriented so that the electrons move horizontally from north to south. At the electron's latitude the vertical component of the Earth's magnetic field points down with a magnitude of 42.3 μT. What is the direction of the force on the electrons due to this component of the magnetic field?

Answers

Answer:

The direction is due south

Explanation:

From the question we are told that

The energy of the electron is $E = 19.0keV = 19.0 *10^3 eV$

The earths magnetic field is $B = 42.3 \muT = 42.3 *10^(-6) T$

Generally the force on the electron is perpendicular to the velocity of the elecrton and the magnetic field and this is mathematically reresented as

$\= F = q (\= v * \=B)$

On the first uploaded image is an illustration of the movement of the electron

Looking at the diagram we can see that in terms of direction the magnetic force is

$\= F =q(\=v * \= B)= -( -\r i * - \r k)$

$= -(- (\r i * \r k))$

generally i cross k = -j

so the equation above becomes

$\= F = -(-(- \r j))$

$= - \r j$

This show that the direction is towards the south

Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s. How much time does the driver of the car measure for his trip between the poles?

Answers

Answer:

Observed time, t = 5.58 s

Explanation:

Given that,

Speed of light in a vacuum has the hypothetical value of, c = 18 m/s

Speed of car, v = 14 m/s along a straight road.

A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s.

We need to find the time the driver of the car measure for his trip between the poles. The relation between real and observed time is given by :

$T=\frac{t}{\sqrt{1-(v^2)/(c^2)} }$

t is observed time.

$t=T* \sqrt{1-(v^2)/(c^2)} \n\nt=8.89* \sqrt{1-(14^2)/(18^2)} \n\nt=5.58\ s$

So, the time observed by the driver of the car measure for his trip between the poles is 5.58 seconds.

The girlfriend of a PHS1282 student proposes to him and offers him a "golden" ring in the process. However, his girlfriend is a student in the Department of Accountancy and Finance, and would struggle to afford a gold ring. Additionally she is known to have a slightly dodgy character. After initially being flattered by the marriage proposal he therefore gets suspicious about whether the ring is actually made of gold. He therefore bring the ring to the PHS「282 lab and measure its mass to be 2.80± 0.02 g and its volume to be 0.16 ± 0.03 cm3 . Gold has a mass density of 19.3 g/cm3. Could the ring be made of gold? (Explain your answer) . Brass has a mass density of 8.4 g/cm3 to 8.7 g/cm3 (depending on the composition of the alloy). Could the ring be made of brass?

Answers

Answer:

The density of the ring is:

$\rho=17.5\pm 3 \, g/cm^3$

This means the ring could very well be made of gold, but it is very unlikely that it is made of brass.

Explanation:

For a quantity f(x,y) that depends on other quantities (in this case two) x and y, the error is given by:

$\sigma_f=\sqrt{\left((\partial f)/(\partial x)\right)^2\sigma_x^2+\left((\partial f)/(\partial y)\right)^2\sigma_y^2 }$

where $\sigma_x$ and $\sigma_y$ are the standard deviations on errors of the variables $x$ and $y$ .

In our case $\rho=f(m,V)=(m)/(V)$ where $m$ is the mass and $V$ is the volume.

Knowing that $\sigma_m=0.02$ and $\sigma_V=0.03$ we can estimate the error on the density

$\sigma_(\rho)=\sqrt{\left((1)/(V)\left)^2\sigma_m^2+\left((m)/(V^2)\right)^2\sigma_V^2}$ $\approx 3$ (values were directly plugged)

The density is by using the given values

$\rho=(m)/(V)=(2.80)/(0.16)=17.5 \, g/cm^3$

The density with error is given by

$\rho=(m)/(V)\pm \sigma_(\rho)=17.5\pm 3 \, g/cm^3$

Which means it could go as high as 20.5 or as low as 14.5, Meaning that the ring could very well be made of gold, but it is very unlikely that it is made of brass.

An 800 kHz radio signal is detected at a point 2.1 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 800 mV/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. The intensity of the radio signal at that point is closest to

Answers

Answer:

$I=8.48* 10^(-4)\ W/m^2$

Explanation:

Given that,

Frequency of the radio signal, $f=800\ kHz=8* 10^5\ Hz$

It is detected at a pint 2.1 km from the transmitter tower, x = 2.1 km

The amplitude of the electric field is, E = 800 mV/m

Let I is the intensity of the radio signal at that point. Mathematically, it is given by :

$I=(E^2_(rms))/(c\mu_o)$

$E_(rms)$ is the rms value of electric field, $E_(rms)=(E)/(√(2) )$

$I=(E^2)/(2c\mu_o)$

$I=((800* 10^(-3))^2)/(2* 3* 10^8* 4\pi * 10^(-7))$

$I=8.48* 10^(-4)\ W/m^2$

So, the intensity of the radio signal at that point is $8.48* 10^(-4)\ W/m^2$ . Hence, this is the required solution.

A 62.0 kg skier is moving at 6.90 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high.How fast is the skier moving when she gets to the bottom ofthe hill?

Answers

Explanation:

The given data is as follows.

Mass, m = 62 kg, Initial speed, $v_(1)$ = 6.90 m/s

Length of rough patch, L = 4.50 m, coefficient of friction, $\mu_(k)$ = 0.3

Height of inclined plane, h = 2.50 m

According to energy conservation equation,

(Final kinetic energy) + (Final potential energy) = Initial kinetic energy + Initial potential energy - work done by the friction

$K.E_(2) + U_(2) = K.E_(1) + U_(1) - W_(f)$

$(1)/(2)mv^(2)_(2) + U_(2) = (1)/(2)mv^(2)_(1) + mgh - \mu_(k)mgL$

Since, final potential energy is equal to zero. Therefore, the equation will be as follows.

$(1)/(2)mv^(2)_(2) = (1)/(2)mv^(2)_(1) + mgh - \mu_(k)mgL$

Cancelling the common terms in the above equation, we get

$(1)/(2)v^(2)_(2) = (1)/(2)v^(2)_(1) + gh - \mu_(k)gL$

= $(1)/(2)(6.90)^(2)_(1) + 9.8 m/s^(2) * 2.50 m - (0.3 * 9.8 * 4.50 m)$

= 36.055 - 13.23

= 22.825

$v_(2) = √(2 * 22.825)$

= 6.75 m/s

Thus, we can conclude that the skier is moving at a speed of 6.75 m/s when she gets to the bottom of the hill.

Answer:

Explanation:

mass, m = 62 kg

initial velocity, u = 6.9 m/s

length, l = 4.5 m

height, h = 2.5 m

coefficient of friction, μ = 0.3

Final kinetic energy + final potential energy = initial kinetic energy + initial potential energy + wok done by friction

Let the final velocity is v.

0.5 mv² + 0 = 0.5 mu² + μmgl + mgh

0.5 v² = 0.5 x 6.9 x 6.9 + 0.3 x 9.8 x 4.5 + 9.8 x 2.5

0.5 v² = 23.805 + 13.23 + 24.5

v² = 123.07

v = 11.1 m/s

Dogs can hear higher-pitched whistles that humans do. How do you
think the sound frequencies that dogs can
hear compare to the frequencies that humans
can hear?

Answers

Dogs can hear sounds at higher frequencies than humans. The range of sound frequencies that dogs can hear is approximately 40 Hz to 60,000 Hz, while the range for humans is 20 Hz to 20,000 Hz. This means that dogs can hear ultrasonic sounds that are beyond the range of human hearing.

What is sound about?

In terms of physics, sound is a vibration that travels through a transmission medium like a gas, liquid, or solid as an acoustic wave.

Sound is the reception of these waves and the brain's perception of them in terms of human physiology and psychology. Dogs have the ability to hear ultrasonic sounds that are audible only to them.

Learn more about sound on:

brainly.com/question/16093793

#SPJ1

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Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979.(a) He flew for 169 min at an average velocity of 3.53 m/s in a direction 45° south of east. What was his total displacement?(b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air?(c) What was his total displacement relative to the air mass?

A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.(a) Find the electric force on charge q2.F12 = ? mN(b) Find the electric force on q1.F21 = ? mN(c) What would your answers for Parts (a) and (b) differ if q2 were -6.0 µC?