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Sedbober [7]
3 years ago
11

The north and south poles of a magnetic field produced by an electromagnet will switch when the direction of the changes.

Physics
2 answers:
baherus [9]3 years ago
4 0

Explanation :

An electromagnet has a coil of wire which is wrapped around an iron core. When an electric current is passed over it, it behaves like a magnet.

For DC current,

When the direction of current is reversed in an electromagnet, its poles gets reversed.

For AC current,

AC or alternating current continuously changes the direction of the current. So, the north poles and south poles keep on changing.

So, the north and south poles of a magnetic field produced by an electromagnet will switch when the direction of the current changes.

Aneli [31]3 years ago
3 0
When the direction of the current changes
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What is the average translational KE of 5 moles of gas molecules at 300 K?
aev [14]

Answer:

What is the average translational kinetic energy of molecules in an ideal gas at 37°C? The average translational energy of a molecule is given by the equipartition theorem as, E = 3kT 2 where k is the Boltzmann constant and T is the absolute temperature.

Explanation:

The average translational energy of a molecule is given by the equipartition theorem as, E = 3kT 2 where k is the Boltzmann constant and T is the absolute temperature.

7 0
3 years ago
What is the relationship between wavelength and the amount of energy the wave carries?
nadezda [96]

Answer:

Energy is inversely proportional to wavelength.

Explanation:

The amount of energy, E, a wave carries is given as:

E = hf

where h = Planck's constant and f = frequency of the wave

Frequency and wavelength are related by the equation:

c = λf

=> f = c/λ

where λ = wavelength

Therefore, energy is:

E = hc/λ

This shows that energy is inversely proportional to wavelength. As wavelength increase, energy decreases and vice versa.

5 0
4 years ago
Numerical Problems
Evgesh-ka [11]

Answer:

a = 5 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f}=v_{o}+a*t

where:

Vf = final velocity = 20 [m/s]

Vo = initial velocity = 10 [m/s]

t = time = 2 [s]

a = acceleration [m/s²]

Now replacing:

20 =10 +a*2\\10=2*a\\a=5[m/s^{2} ]

4 0
3 years ago
A spaceship starting from a resting position accelerates at a constant rate of 9.8 meters per second per second. How long and ho
Verizon [17]

Accelerating at 9.8 m/s² means that every second, the speed is 9.8 m/s faster than it was a second earlier.  It's not important to the problem, but this number (9.8) happens to be the acceleration of gravity on Earth.

1% of the speed of light = (300,000,000 m/s) / 100 = 3,000,000 m/s .

Starting from zero speed, moving (9.8 m/s) faster every second,
how long does it take to reach  3,000,000 m/s ?

           (3,000,000 m/s) / (9.8 m/s²)  =  306,122 seconds .
                                                   (That's  5,102 minutes.)
                                                        (That's  85 hours.)
                                                     (That's  3.54 days.)

Speed at the beginning . . . zero .
Speed at the end . . . 3,000,000 m/s
Average speed . . . . . 1,500,000 m/s

Distance = (average speed) x (time)

               = (1,500,000 m/s) x (306,122 sec) = 4.592 x 10¹¹ meters

                                                                     =  459 million kilometers

                         That's like from Earth
                                                  to       Sun
                                                             to    Earth
                                                                    to        Sun. 

4 0
3 years ago
A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk
Murrr4er [49]

Answer:

\omega_2=5.1rad/s

Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is L=I\omega. If we call 1 the situation when the student is at the rim and 2 the situation when the student is at r_2=1.39m from the center, then we have:

L_1=L_2

Or:

I_1\omega_1=I_2\omega_2

And we want to calculate:

\omega_2=\frac{I_1\omega_1}{I_2}

The total moment of inertia will be the sum of the moment of intertia of the disk of mass m_D=119.1 kg and radius r_D=3.23m, which is I_D=\frac{m_Dr_D^2}{2}, and the moment of intertia of the student of mass m_S=54.3kg at position r (which will be r_1=r=3.23m or r_2=1.39m) will be I_{S}=m_Sr_S^2, so we will have:

\omega_2=\frac{(I_D+I_{S1})\omega_1}{(I_D+I_{S2})}

or:

\omega_2=\frac{(\frac{m_Dr_D^2}{2}+m_Sr_{S1}^2)\omega_1}{(\frac{m_Dr_D^2}{2}+m_Sr_{S2}^2)}

which for our values is:

\omega_2=\frac{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(3.23m)^2)(3.1rad/s)}{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(1.39m)^2)}=5.1rad/s

6 0
3 years ago
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