Continental drift is_____

Match these items. 1. Ca 6 proton 2. H 2O 3 fission 3. nuclear decay 7 element 4. nuclear synthesis 8 electron 5. η atomic numbe

Gnesinka [82]

1. Ca → Element

2. Proton → positive

3. H2O → compound

4. Fission → nuclear decay

5. Fusion → Nuclear synthesis

6. η → Neutron

7. e → electron

8. Atomic number → no of protons in nucleus.

Explanation

1. Ca (Calcium):

Calcium is an element with the atomic number of 20. It is an alkaline earth metal. The 99% of calcium is found in our bodies, in bones, teeth.

2. Proton:

Proton is a subatomic particle and it holds the positive charge. Proton is present in the nucleus of the atom.

3. H2O (water):

Water is a chemical compound and it's chemical formula is H2O. It's called compound as it contains 2 hydrogen and 1 oxygen atoms bonded together through the covalent bond.

4. Fission:

Fission is a process in which large massive unstable nucleus splits into the smaller, less heavier and stable nuclei. The energy is re;eased in the form of radiations during this process. It's called as the radioactive decay.

5. Fusion:

Fusion is opposite of the fission reaction. As in this case the two nuclei combines to form a single large nucleus. That's why it is a nuclear synthesis process.

6. η neutron:

Neutron is a subatomic particle and it is a neutral particle which is located inside the nucleus. n is a symbol used for the neutron.

7. e Electron:

The symbol for electron is e. It's a subatomic particle with negative charge. It is found in the orbits around the nucleus.

8. Atomic Number:

Atomic number is defined as the number of protons in the nucleus of an atom. IT is represented by Z.

6 0

3 years ago

Positive feedback interactions in earth’s systems are always a result of human action. T/F

VLD [36.1K]

Positive feedback interactions in earth’s systems are always a result of human action is a FALSE statement.

<u>Explanation:</u>

Earth is a unstable equilibrium which tends to move out of equilibrium, but several other factors try to bring it back in equilibrium again. Earth has a different actions going on both on its surface and also inside it.

Human can alter, or can modify a very small part of the events that occur on earth’s surface. But they don’t have any control on what’s going inside earth’s core. So positive feedback interactions are not only the results of human interaction, but also different other factors.

4 0

3 years ago

Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad

love history [14]

Answer:

a) The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ , b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

$\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where:

$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

$\alpha$ - Angular acceleration, measured in radians per square second.

$r$ - Radius of rotation (Radius of a tire), measured in meters.

Given that $\alpha = 15\,\frac{rad}{s^{2}}$ and $r = 0.31\,m$ . The linear acceleration experimented by the car is:

$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

$\theta = 46.875\,rad$

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0

3 years ago

Whats the answer to this question show in the picture 2 questions

astraxan [27]

1) D

2) I would say A, but not 100%, its the only one that makes sense tho

3 0

3 years ago

Read 2 more answers

Explain why models are often used to study other planets ?

bonufazy [111]

The models are used to represent what you are studying in this case would be a planet. A model of Saturn and its rings and the moons surrounding it would be fantastic to look at when you have no way of going there

7 0

3 years ago

Continental drift is______?