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3 years ago

How are solar flares different from solar prominences.

2 answers:

4 0

A prominence is a loop of cool incandescent gas that extends above the photosphere. A solar flare is an explosive release of energy that comes from the sun and causes magnetic ditrubances.

nadya68 [22]3 years ago

4 0

Answer:

What is the difference between a prominence and a solar flare? A prominence is a loop of cool incandescent gas that extends above the photosphere. A solar flare is an explosive release of energy that comes from the sun and causes magnetic ditrubances.

Explanation:

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Magnesium (Mg) has two electrons in its outermost energy level. Oxygen (O) has six. How can a Mg atom body with an O atom?

Rudiy27

I’ll take the points

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3 years ago

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A car's position in relation to time is plotted on the graph. What can be said about the car during segment B?

CaHeK987 [17]

we know that

the speed is equal to

$speed=\frac{distance}{time}$

The slope of the line on the graph is equal to the speed of the car

during the segment B the slope of the line is equal to zero

that means

the speed of the car is zero

therefore

<u>the answer is the option B</u>

The car has come to a stop and has zero velocity

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3 years ago

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A system of two objects has ΔKtot = 6 J and ΔUint = -5 J. Part A How much work is done by interaction forces? Express your answe

Elina [12.6K]

A) +5 J

B) +1 J

Explanation:

The internal forces (interaction forces) acting on a system do not change the mechanical energy (sum of potential and kinetic energy) of the system.

However, these forces are responsible for converting the energy from one form into another; the work done by these forces is equal to the amount of energy converted from one form into the other.

In this problem, we have:

$\Delta U=-5 J$ is the loss in potential energy of the system

$\Delta K=+6 J$ is the gain in kinetic energy of the system

By looking at these numbers, this means that the internal forces have converted 5 J of energy from potential energy into kinetic energy (while the additional +1 J missing is due to external forces, as explained in part B).

Therefore, the work done by internal forces is

W = +5 J

First of all, we calculate the change in mechanical energy of the system.

The mechanical energy of a system is the sum of its kinetic energy (K) and its potential energy (U):

$E=K+U$

So, the change in mechanical energy is equal to the sum of the changes of kinetic energy and the changes of potential energy:

$\Delta E= \Delta K + \Delta U$

In this problem:

$\Delta K=+6 J$

$\Delta U=-5 J$

So, the change in mechanical energy is:

$\Delta E=+6+(-5)=+1 J$

According to the work-energy theorem, the work done by external forces on a system is equal to the change in mechanical energy of the system: therefore in this case, the work done by external forces is

$W=\Delta E=+1 J$

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4 years ago

A baseball team is practicing throwing balls vertically upward to test their throwing arms. A player manages to throw a ball tha

Nitella [24]

We can solve this problem using the law of conservation of energy.
This law states that energy in a closed system must stay same.
That means that the energy of a ball leaving the hand and the energy of a ball when it reaches its maximum height must be the same.
The energy of a ball leaving the players hand is kinetic energy:
$E_k=\frac{mv^2}{2}$
The energy when the ball reaches its maximum height ( and has zero velocity) is potential energy in a gravitational field:
$E_p=mgh$
As said before these energies must be the same, and that allows us to find the initial speed:
$E_k=E_p\\
\frac{mv^2}{2}=mgh\\
v^2=2gh\\
v=\sqrt{2gh}$
When we plug in all the number we get that $v=14\frac{m}{s}$

7 0

3 years ago

A dockworker applies a constant horizontal force of 73.0 N to a block of ice on a smooth horizontal floor. The frictional force

vivado [14]

Answer:

a) 57.0 kg b) 24.2 m

Explanation:

a) According Newton's second law, the applied force is equal to the product of the mass times the acceleration.

As the force is constant, the acceleration is constant too.

In this case, as we have as givens the distance and the time, and also we know that the block is starting form rest, we can get the acceleration as follows:

d = 1/2 * a * t² ⇒ a = 2d / t² ⇒ a= 2* 13.0 m / (4.5)² s² = 1.28 m/s²

Replacing in the Newton's 2nd Law equation:

F = m*a ⇒ m = F/a = 73.0 N / 1.28 m/s = 57.0 Kg

b) At t=4.5 s, applying the definition of acceleration, we can get the value of the velocity at that time, as follows:

v= a* t = 1.28 m/s * 4.5 s = 5.76 m/s

If the worker stops pushing at the end of the 4. 5 s, this means (neglecting friction) that from that time omwards, no net force acts on the block, so it continues moving at constant speed.

In order to get the distance moved in the next 4.20 sec, as it is moving at constant speed, we neeed just to apply the definition of velocity:

v= Δx / Δt ⇒ Δx = v* Δt = 5.76 m/s * 4.2 m = 24.2 m

So, the total distance traveled during all the time (9.1 s) is just the sum of the 13.0 m advanced during the time when there was a constant force applied, and the last 24.2 m at constant speed, as follows:

d = 13.0 + 24.2 = 37.2 m

3 0

3 years ago