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3 years ago

When do sunspots disappear?

Physics

1 answer:

dedylja [7]3 years ago

8 0

Answer:

In 5 years or so, the sun will be awash in sunspots and more prone to violent bursts of magnetic activity.

Explanation

once the magnetic field weakens the area and cold plasma enters the area of the sunspot

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When a pendulum is at the position all the way to the left when it is swinging (at the top of the arc), what is true of the kine

JulsSmile [24]

Choice - B is the correct one.

At the top of the arc, at one end of the swing:
-- it's not going to get any higher, so the potential energy is maximum
-- it stops moving for an instant, so the kinetic energy is zero

At the bottom of the arc, in the center of the swing:
-- it's not going to get any lower, so the potential energy is minimum
-- it's not going to move any faster, so the kinetic energy is maximum

7 0

3 years ago

A mass is oscillating with amplitude A at the end of a spring.

Dmitry_Shevchenko [17]

A) $x=\pm \frac{A}{2\sqrt{2}}$

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

$E=\frac{1}{2}kA^2$ (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

$U=\frac{1}{3}K$

Using (2) we can rewrite this as

$U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}$

And using (1), we find

$U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2$

Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

$\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}$

B) $x=\pm \frac{3}{\sqrt{10}}A$

In this case, the kinetic energy is 1/10 of the total energy:

$K=\frac{1}{10}E$

Since we have

$K=E-U$

we can write

$E-U=\frac{1}{10}E\\U=\frac{9}{10}E$

And so we find:

$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

3 0

3 years ago

An 4-kg ball experiences a force and accelerates at a rate of 1.5 m/s.

ANTONII [103]

Answer:

6.0 N

Explanation:

The strength of a force is expressed as the magnitude of the force in Newton.

The formula to apply here is :

Force= mass * acceleration

F=ma

Mass, m = 4 kg

Acceleration = 1.5 m/s²

Force= 4 *1.5 = 6.0 N

4 0

3 years ago

Which inference about Arthur is supported by the text?

yarga [219]

There is no picture

8 0

3 years ago

The first and second coils have the same length, and the third and fourth coils have the same length. They differ only in the cr

tatyana61 [14]

Answer:

The ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.

And

The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.

Explanation:

The resistance of the coil is directly proportional to the length of the coil and inversely proportional to the area of coil and hence inversely proportional to the square of radius of the coil.

So, the ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.

And

The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.

8 0

3 years ago