What is the meaning (definition) of the magnet forces, electric forces, electromagnetism

A 2.5 kg block is launched along the ground by a spring with a spring constant of 56 N/m. The spring is initially compressed 0.7

Norma-Jean [14]

vf ^2 = kx^2/m = 56(0.75)^2 / 2.5 = 12.6

Therefore, v= 3.5 m/s.

7 0

3 years ago

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Most of the really bright stars in our sky are NOT among the stars that are very close to us. Why then do they look so bright to

Yanka [14]

Answer:

stars will emit more light due to their Luminosity, so they look very bright.

Explanation:

Luminous refers to..,

The total amount of energy radiated by a star or other celestial object per second.
Therefore it is the power output of a star.

Most of the really bright stars in our sky are not that very close to us yet they look bright because of the Luminosity of the star.

These stars are intrinsically so luminous.

A star's power output across all wavelengths is called its bolometric luminosity.

A star with large luminosity will have more measure of radiated electromagnetic power meaning.

so it will emit more light than a low luminosity star.

Hence,

those stars can easily be seen even across great distance.

learn more about Luminosity of the star here:

<u>brainly.com/question/13912549</u>

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6 0

2 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0

3 years ago

A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a

sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0

3 years ago

during conversion water to ice,forces of attraction between molecules. OPTIONS:1.)Decreases 2.)Increases 3.)Does not change 4.)C

In-s [12.5K]

because water is loosely packed but when it is cold it becomes closely packed in order to form ice and thus the force attraction between them also increase.

4 0

3 years ago