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3 years ago

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A long, straight solenoid has 800 turns. when the current in the solenoid is 2.90 a, the average flux through each turn of the s

olenoid is 3.25 * 10-3 wb. what must be the magnitude of the rate of change of the current in order for the self-induced emf to equal 6.20 mv?

1 answer:

tresset_1 [31]3 years ago

4 0

It has to be the right answer for it to be correct

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Latitude and longitude are the two coordinates that determine a specific point on the Earth’s surface. How does knowing the loca

DedPeter [7]

Answer:

hey sandra! i hope this will help you!

Explanation:

The geographic coordinate system is a reference system that uses the two angular latitude (north or south) and longitude (east or west) coordinates to determine the positions of the land surface points. These two angular coordinates measured from the center of the Earth are from a spherical coordinate system that is aligned with its axis of rotation. These coordinates are usually expressed in sexagesimal degrees:

Latitude measures the angle between any point and the equator. Latitude lines are called parallel and are circles parallel to the equator on the Earth's surface. Latitude is the distance between any point and Ecuador, measured on the meridian that passes through that point.

• All points located on the same parallel have the same latitude.

• Those who are in the north of Ecuador receive the denomination North (N).

• Those who are in the south of Ecuador receive the denomination South (S).

• It is measured from 0º to 90º.

• The latitude of 0º corresponds to Ecuador.

• The North and South poles have latitude 90º N and 90º S respectively.

The length measures the angle along the equator from any point on Earth. It is accepted that Greenwich in London is length 0 in most modern societies. The lines of length are maximum circles that pass through the poles and are called meridians.

• All points located on the same meridian have the same length.

• Those who are east of Meridian Zero receive the designation East (E).

• Those who are west of Meridian Zero receive the denomination West (O).

• It is measured from 0º to 180º.

• The Greenwich meridian corresponds to the length 0º.

By combining these two angles, the position of any point on the Earth's surface can be expressed.

6 0

2 years ago

Plz help me answer this is my Final Exam!!!

Ostrovityanka [42]

It's answer number four

8 0

3 years ago

Ice floats on water.

OleMash [197]

Answer:

Yeah ice floats on water.

Observation

Example in those areas were ice is found like Antarctica ice is found on top of water.

5 0

2 years ago

Determine which heat transfers below are due to the process of conduction. I) You walk barefoot on the hot street and it burns y

Taya2010 [7]

Answer:

I) You walk barefoot on the hot street and it burns your toes.

II) When you get into a car with hot black leather in the middle of the summer and your skin starts to get burned.

Explanation:

In conduction mode of heat transfer we know that the energy is transferred from one system to other system due to direct contact of two bodies

Here due to this direct contact the energy is transferred via a given solid or liquid medium

In this type of heat transfer medium particles will remain in its own position only the energy is transferred.

So here we can say the correct answer will be

I) You walk barefoot on the hot street and it burns your toes.

II) When you get into a car with hot black leather in the middle of the summer and your skin starts to get burned.

3 0

3 years ago

A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput

Nikitich [7]

There is a missing data in the text of the problem (found on internet):
"with force constant<span> k=</span>450N/<span>m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
</span> $E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
<span>where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
</span> $E=K_{max} = \frac{1}{2}mv_{max}^2$
<span>Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
</span> $E=U_{max}= \frac{1}{2}k A^2$
<span>
Since the mechanical energy must be conserved, we can write
</span> $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
<span>from which we find the maximum speed
</span> $v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$
<span>
b) </span><span> the </span>speed<span> of the </span>glider<span> when it is at x= -0.015</span><span>m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
</span> $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$
<span>
At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
</span> $U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
<span>And since
</span> $E=U+K$
<span>we find the kinetic energy when the glider is at this position:
</span> $K=E-U=0.36 J - 0.05 J = 0.31 J$
<span>And then we can find the corresponding velocity:
</span> $K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$
<span>
c) </span><span>the magnitude of the maximum acceleration of the glider;
</span>
For a simple harmonic motion, the magnitude of the maximum acceleration is given by
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) <span>the </span>acceleration<span> of the </span>glider<span> at x= -0.015</span><span>m

For a simple harmonic motion, the acceleration is given by
</span> $a(t)=\omega^2 x(t)$
<span>where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
</span> $a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$
<span>
e) </span><span>the total mechanical energy of the glider at any point in its motion. </span><span>

we have already calculated it at point b), and it is given by
</span> $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

8 0

3 years ago