Egocentrism, animism, and artificialism are characteristic of which of Jean Piaget's stages of cognitive development?

What is the most likely radiological device that a terrorist might use in a Weapons of Mass Destruction (WMD) incident?

zavuch27 [327]

Answer:Radiologic Dispersal Devices (RDD)

Explanation:

Radiologic Dispersal Device (RDD) is commonly known as dirty bomb. These devices require little more skill than is needed to make a conventional bomb and their components are easier to acquire. RDDs utilize conventional explosives to disperse a radioactive material packaged in the device, as opposed to a nuclear device, which creates radiation with its explosion.

7 0

2 years ago

Which energy resources are found above the Earth's surface

never [62]

Answer: the sun

Explanation:

The sun's radiant energy reaches the earth's surface either directly through radiation, indirectly through convection, or it can move "across" or "through" objects or materials on the surface via conduction. Let's look more closely at each case. We've probably experienced the feeling of "warmth" of the sun on our skin on a sunny day. Light energy from the sun is reaching us across space and down through the atmosphere through radiation. A dark colored vehicle in the sun quickly becomes warm (or hot!) to the touch because of radiation. The light energy from the sun heats the air in the earth's atmosphere, and this drives convection and transfers thermal energy around. It is possible that we've felt a "hot breeze" on our skin on sunny days. The thermal energy in the air will be carried to objects in its path, and it will warm them.

6 0

2 years ago

A block rides on a piston that is moving vertically with simple harmonic motion. (a) If the SHM has period 2.68 s, at what ampli

fredd [130]

Answer:

Part a)

$A = 1.78 m$

Part b)

$f = 2 rev/s$

Explanation:

Part A)

As we know that time period of the motion is given as

$T = 2.68 s$

so we have

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

$\omega = \frac{2\pi}{2.68}$

$\omega = 2.34 rad/s$

now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

$mg = m\omega^2 A$

so we have

$A = \frac{g}{\omega^2}$

$A = \frac{9.81}{2.34^2}$

$A = 1.78 m$

Part b)

Now if the amplitude of the SHM is 6.23 cm

and now at this amplitude if object will lose the contact then in that case again we have

$mg = m\omega^2 A$

$g = \omega^2 (0.0623)$

$\omega = 12.5 rad/s$

so now we have

$2\pi f = 12.5$

$f = 2 rev/s$

3 0

2 years ago

Kinematics

leonid [27]

Answer:

a)

a = 2 [m/s^2]

b)

a = 1.6 [m/s^2]

c)

xt = 2100 [m]

Explanation:

In order to solve this problem we must use kinematics equations. But first we must identify what kind of movement is being studied.

a)

When the car moves from rest to 40 [m/s] by 20 [s], it has a uniformly accelerated movement, in this way we can calculate the acceleration by means of the following equation:

$v_{f} = v_{i}+(a*t)$

where:

Vf = final velocity = 40 [m/s]

Vi = initial velocity = 0 (starting from rest)

a = acceleration [m/s^2]

t = time = 20 [s]

40 = 0 + (a*20)

a = 2 [m/s^2]

The distance can be calculates as follows:

$v_{f} ^{2} = v_{i} ^{2}+(2*a*x)$

where:

x1 = distance [m]

40^2 = 0 + (2*2*x1)

x1 = 400 [m]

Now the car maintains its speed of 40 [m/s] for 30 seconds, we must calculate the distance x2 by means of the following equation, it is important to emphasize that this movement is at a constant speed.

v = x2/t2

where:

x2 = distance [m]

t2 = 30 [s]

x2 = 40*30

x2 = 1200 [m]

b)

Immediately after a change of speed occurs, such that the previous final speed becomes the initial speed, the new Final speed corresponds to zero, since the car stops completely.

$v_{f} = v_{i}-a*t$

Note: the negative sign of the equation means that the car is stopping, i.e. slowing down.

0 = 40 - (a *25)

a = 40/25

a = 1.6 [m/s^2]

The distance can be calculates as follows:

$v_{f} ^{2} = v_{i} ^{2} -2*a*x3\\$

0 = (40^2) - (2*1.6*x3)

x3 = 500 [m]

c)

Now we sum all the distances calculated:

xt = x1 + x2 + x3

xt = 400 + 1200 + 500

xt = 2100 [m]

8 0

2 years ago

Which takes place as water cycles from the bottom of the pot toward the top?

Elenna [48]

The option that takes place as water cycles from the bottom of the pot toward the top is that A. thermal energy is transferred.
As the pot gets warmer and warmer, the heat flows everywhere inside the pot, ultimately reaching the top, and heating the water at the top as well. There is no chemical energy here, and molecules don't gain thermal energy, it is just transferred to the top of the pot.

8 0

2 years ago

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