The digital exchange of structured data is called ?

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Answer 1

Answer:

Answer:

Electronic data interchange

Related Questions

A dynamics cart with a friction pad is placed at the top of an inclined track and released fromrest. The cart accelerates down the incline at a rate of 0.60 m/s2. If the track is angled at 10degrees above the horizontal, determine the coefficient of kinetic friction between the cart andthe track.
3.what does this stand for ??
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A 75 kg man starts climbing a ladder that leans against a wall. If the weight of the ladder is negligible, determine how far up the ladder the man can climb before the ladders starts to slip. The coefficient of friction on both surfaces is μS=0.25

Look at the figure below and calculate the length of side y.A. 8.5
B. 17
12
y
C. 6
O D. 12
45
Х

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Answer:

I want to say a because you want to subtract and simplify

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in meters, t in seconds, and c and b are positive constants. What are the units of (a) constant c and (b) constant b? Find a formula in terms of c, b, and t of the (c) velocity v and (d) acceleration a. (e) At what time t ≥ 0 does the particle reach its maximum x value?

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Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=(2c)/(3b).$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=(dx)/(dt)\n=(d)/(dx)(ct^2-bt^3)\n=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = (dv)/(dt)\n=(d)/(dt)(2ct-3bt^2)\n=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left ((dx)/(dt)\right )_(t=t_o)=0.$
$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.$

Applying both these conditions,

$\rm \left ( (dx)/(dt)\right )_(t=t_o)=0\n2ct_o-3bt_o^2=0\nt_o(2c-3bt_o)=0\nt_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = (2c)/(3b).$

For $\rm t_o = 0$ ,

$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = (2c)/(3b)$ ,

$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6b\cdot (2c)/(3b)=2c-4c=-2c.$

Here,

$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.$

Thus, the particle reach its maximum x value at time $\rm t_o = (2c)/(3b).$

A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processes are generating, its temperature will drop. If the drop is severe, it can be life-threatening. Suppose that a person is unclothed and energy is being lost via radiation from a body surface area of 1.36 m2, which has a temperature of 34° C and an emissivity of 0.700. Also suppose that metabolic processes are producing energy at a rate of 122 J/s. What is the temperature of the coldest room in which this person could stand and not experience a drop in body temperature

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To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzman law that is responsible for calculating radioactive energy.

Mathematically this expression can be given as

$P = \sigma Ae\Delta T^4$

Where

A = Surface area of the Object

$\sigma =$ Stefan-Boltzmann constant

e = Emissivity

T = Temperature (Kelvin)

Our values are given as

$A = 1.36m^2$

$\Delta T^4 = T_2^4 -T_1^4 = 307^4-T_1^4$

$\sigma = 5.67*10^(-8) J/(s m^2 K^4)$

$P = 122J/s$

$e = 0.7$

Replacing at our equation and solving to find the temperature 1 we have,

$P = \sigma Ae\Delta T^4$

$P = \sigma Ae (T_2^4 -T_1^4)$

$122 = (5.67*10^(-8))(1.36)(0.7)(307^4-T_1^4)$

$T_1 = 285.272K = 12.122\°C$

Therefore the the temperature of the coldest room in which this person could stand and not experience a drop in body temperature is 12°C

For every increase in mass the gravitational force blank If the total mass increase by effective for the gravitational force

Answers

For every increase in mass, the gravitational force increases. Gravitational force is directly proportional to the mass of the object.

What is gravitational force?

Gravitational force is the force by which an object attracts other objects into its center of mass. Earth attracts other objects gravitationally and that keep everyone stand to the ground.

Gravitational force directly proportional to the mass and inversely proportional to the distance between the objects. The expression relating the force and mass is written as:

g = G m/r²

Where G is the universal gravitational constant.

Therefore, as the mass of the object increase, the gravitational force exerted also increases. Similarly massive object experience more gravitation force by earth.

Find more on gravitational force:

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Answer:

Increases by the same amount.Increases by a factor of 4.

Explanation:

i took it

An infant throws 7 g of applesauce at a velocity of 0.5 m/s. All of the applesauce collides with a nearby wall and sticks to it. What is the decrease in kinetic energy of the applesauce?

Answers

Answer:

Δ KE = - 8.75 x 10⁻⁴ J

Explanation:

given,

mass of applesauce = 7 g = 0.007 Kg

initial velocity, u = 0.5 m/s

final velocity, v = 0 m/s

Decrease in kinetic energy = ?

initial kinetic energy

$KE_1=(1)/(2)mu^2$

$KE_1=(1)/(2)* 0.007 * 0.5^2$

KE₁ = 8.75 x 10⁻⁴ J

final kinetic energy

$KE_2=(1)/(2)mv^2$

$KE_2=(1)/(2)* 0.007 * 0^2$

KE₂ =0 J

Decrease in kinetic energy

Δ KE = KE₂ - KE₁

Δ KE = 0 - 8.75 x 10⁻⁴

Δ KE = - 8.75 x 10⁻⁴ J

decrease in kinetic energy of the applesauce is equal to 8.75 x 10⁻⁴ J

Final answer:

The decrease in kinetic energy of the applesauce, when it hits the wall and stops, is the initial kinetic energy of it. Using the formula of kinetic energy, the decrease is calculated to be 0.000875 Joules.

Explanation:

This question relates to the concept of kinetic energy in physics. Kinetic energy is calculated by the formula 0.5 * mass (kg) * velocity (m/s)^2. So the initial kinetic energy of the applesauce right after being thrown was 0.5 * 0.007 kg * (0.5 m/s)^2 = 0.000875 Joules.

When the applesauce hits the wall and stops, its velocity drops to 0. Thus, its kinetic energy also goes to 0 (because kinetic energy is proportional to the square of velocity).

Therefore, the decrease in kinetic energy is the same as the initial kinetic energy of the applesauce, which is 0.000875 Joules.

Learn more about Kinetic Energy here:

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g Adjacent rows in the first part of the experiment are found to have potentials of 3.66 V and 4.22 V. If the distance between rows is found to be 0.4 cm, what is the magnitude of the electric field at the location between the rows

Answers

Answer:

$E=140V/m$

Explanation:

If the electric field is uniform, the electric field between two points at potentials $V_1$ and $V_2$ which are separated by a distance d will be given by the formula:

$E=(\Delta V)/(d)$

So in our case, we have $E=(4.22V-3.66V)/(0.004m)=140V/m$

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Consider different points along one spoke of a wheel rotating with constant angular velocity. Which of the following is true regarding the centripetal acceleration at a particular instant of time?a. The magnitude of the centripetal acceleration is greater for points on the spoke closer to the hub than for points closer to the rimb. both the magnitude and the direction of the centripetal acceleration depend on the location of the point on the spoke.c. The magnitude of the centripetal acceleration is smaller for points on the spoke closer to the hub than for points closer to the rim but the direction of the acceleration is the same at all points on this spoke.d. The magnitude and direction of the centripetal acceleration is the same at all points on this spoke.

1. Explain the change of state from solid dry ice to carbon dioxide gas.2. The motion of the particles in dry ice and carbon dioxide gas.3. Explain how the original mass of dry ice compares with the mass of carbon dioxide gas.

A pilot performs an evasive maneuver by diving vertically at 270. If he can withstand an acceleration of 9.0 's without blacking out, at what altitude must he begin to pull out of the dive to avoid crashing into the sea? y= ?m

What is the weight on Earth of an object with mass 45 kg. Hint gravity = 10 N/kg *1 point45 N450 N450 kg10N