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Mazyrski [523]
3 years ago
14

Which type of atom has the same number of protons and electrons

Physics
1 answer:
beks73 [17]3 years ago
5 0
The answer is hydrogen
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Explain what each quantum number in a quantum number set tells you about the electron. Compare and contrast the locations and pr
Allisa [31]
There are four quantum numbers:
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2) Azimuthal quantum number which tells the subshell of the electron. This has a value of an integer starting from 0, 0 being the s orbital. The first electron is in the d orbital due to the number being 2 and the second is in the p orbital due to the number being 1.

3) Magnetic quantum number tells the orbital within the subshell. The first electron is in the -1 orbital of the d subshell (which has values from -2 to 2) and the second is in the -1 orbital of the p subshell (which has values from -1 to 1).

4) Spin quantum number which specifies the spin on the electron, both of the electrons have the same spin.
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3 years ago
Why does the density of a substance remain the same for different amount of the substance
tresset_1 [31]

Think of it this way: 
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-- If a certain salad dressing has 40 calories PER Tablespoon, it doesn't
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6 0
3 years ago
Which best explains why the receiver of a signal must understand the code or language being used
Oksi-84 [34.3K]

The receiver of a signal must understand the code or language being used to avoid confusion and losses.

<h3>What is a Signal?</h3>

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4 0
2 years ago
What mechanism is most responsible for generating the internal heat of Io that drives its volcanic activity?
Ghella [55]

Answer:

Tidal heating

Explanation:

Tidal force is the ability of a massive body to produce tides on another body. The tidal force depends on the mass of the body that produces the tides and the distance between the two bodies.

Tidal forces can cause the destruction of a satellite that orbits a planet or a comet that is too close to the Sun or a planet. When the orbiting body crosses the "Roche boundary", the tidal forces along the body are more intense than the cohesion forces that hold the body together.

Tidal friction is the force between the Earth's oceans and ocean floors caused by the gravitational attraction of the Moon. The Earth tries to transport the waters of the oceans with it, while the Moon tries to keep them under it and on the opposite side of the Earth. In the long term, tidal friction causes the Earth's rotation speed to decrease, thus shortening the day. In turn, the Moon increases its angular momentum and gradually spirals away from Earth. Finally, when the day equals the orbital period of the Moon (which will be about 40 times the length of the current day), the process will cease. Subsequently, a new process will begin when the power to raise tides from the Sun takes angular momentum from the Earth-Moon system. The Moon will then spiral towards Earth until it is destroyed when it enters the "Roche boundary."

<u>Tidal heating </u>

It is the warming caused by the tidal action on a planet or satellite. The most important example of tidal heating in the Solar System is the effect of Jupiter on its Io satellite, in which the tidal effects produce such high temperatures that the interior of the satellite melts, producing volcanism.

8 0
3 years ago
A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati
Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

  • If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

       \tau = I * \alpha  (1)

  • Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
  • Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
  • τ = F*r (2)
  • For a solid uniform disk, the rotational inertia regarding an axle passing through its center  is just I = m*r²/2 (3).
  • Replacing (2) and (3) in (1), we can solve for α, as follows:

       \alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)

  • Since the angular acceleration is constant, we can use the following kinematic equation:

        \omega_{f}^{2}  - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)

  • Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

       0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)

  • Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

       \omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)

  • Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

        v = \omega * r (8)

  • where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of  the disk, r becomes the radius of the disk, 0.200 m.
  • Replacing this value and (7) in (8), we get:

       v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)

b)    

  • There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

       a_{t} = \alpha * r (9)

  • where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
  • Replacing this value and (4), in (9), we get:

       a_{t}  = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)

  • Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
  • The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

       a_{c} = \omega^{2} * r  (11)

  • Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
  • Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
  • Replacing this value and (7) in (11) we get:

       a_{c} = \omega^{2} * r   = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)

  • The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
  • Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

       a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)

6 0
3 years ago
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