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iogann1982 [59]
3 years ago
13

Please select the word from the list that best fits the definition. The movement of a magnetic pole away from the actual pole

Physics
1 answer:
elena55 [62]3 years ago
7 0

Answer:

Magnetosphere

Explanation:

It's the magnetosphere simply because a magnetic declination is not on abject, it's something that happens.

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Imagine an alternate universe where all of the quantum number rules were identical to ours except m_{s} had three allowed values
marishachu [46]

Answer:

so in a given orbital there can be 3 electrons.

Explanation:

The Pauli exclusion principle states that all the quantum numbers of an electron cannot be equal, if the spatial part of the wave function is the same, the spin part of the wave function determines how many electrons fit in each orbital.

In the case of having two values, two electrons change. In the case of three allowed values, one electron fits for each value, so in a given orbital there can be 3 electrons.

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3 years ago
What are northern lights? a lights on mars b lights on the sun c lights on venus
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The northern lights are shafts or curtains of colorful light that occasionally appear in the night sky. They are one of the numerous astronomical phenomena known as polar lights (aurora Polaris).This phenomenon may be observed in mars.

Earth's magnetic field directs electrons and protons from the sun to the poles, where they excite atmospheric gas molecules and cause them to glow, resulting in the aurora borealis and aurora australis, two nocturnal light displays. You might refer to it as the aurora Universalis on Mars. This is because Mars does not direct the energetic particles from the sun to its poles since it lacks an internal magnetic field. Today, researchers utilizing the MAVEN (Mars Atmosphere and Volatile Evolution) spacecraft find evidence for an aurora that may potentially cover the whole nightside of the planet. Venus lacks a magnetic field, thus it would not experience the same kind of nighttime aurora that we do.

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8 0
2 years ago
uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station
Alex787 [66]

Answer:

The magnitude of the tangential velocity is v= 0.868 m/s

The magnitude of the resultant acceleration at that point is  a = 4.057 m/s^2

Explanation:

From the question we are told that

      The mass of the uniform disk is m_d = 40.0kg

       The radius of the uniform disk is R_d = 0.200m

       The force applied on the disk is F_d = 30.0N

Generally the angular speed i mathematically represented as

             w = \sqrt{2 \alpha  \theta}

Where \theta is the angular displacement given from the question as

           \theta  = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}

                 =1.257\  rad

   \alpha is the angular acceleration which is mathematically represented as

                    \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

    The moment of inertial is mathematically represented as

                     I = \frac{1}{2} m_dR^2_d

Substituting values

                    I = 0.5 * 40 * 0.200^2

                        = 0.8kg \cdot m^2

Considering the equation for angular acceleration

               \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

Substituting values

               \alph\alpha = \frac{(30.0)(0.200)}{0.8}

                   = 7.5 rad/s^2

Considering the equation for angular velocity

    w = \sqrt{2 \alpha  \theta}

Substituting values

     w =\sqrt{2 * (7.5) * 1.257}

         = 4.34 \ rad/s

The tangential velocity of a given point on the rim is mathematically represented as

                 v = R_d w

Substituting values

                    = (0.200)(4.34)

                     v= 0.868 m/s

The radial acceleration at hat point  is mathematically represented as

            \alpha_r = \frac{v^2}{R}

                  = \frac{0.868^2}{0.200^2}

                 = 3.7699 \ m/s^2

The tangential acceleration at that point is mathematically represented as

               \alpha _t = R \alpha

Substituting values

           \alpha _t = (0.200) (7.5)

                 = 1.5 m/s^2

The magnitude of resultant acceleration at that point is

                 a = \sqrt{\alpha_r ^2+ \alpha_t^2 }

Substituting values

                a = \sqrt{(3.7699)^2 + (1.5)^2}

                   a = 4.057 m/s^2

         

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Differentiate between gravity and<br>gravitation<br>​
ZanzabumX [31]
Gravitation is the attractive force existing between any two objects that have mass. The force of gravitation pulls objects together. Gravity is the gravitational force that occurs between the earth and other bodies. Gravity is the force acting to pull objects toward the earth.
6 0
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