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3 years ago

What is the definition of energy transformation?

Physics

2 answers:

mixas84 [53]3 years ago

4 0

Energy Transformation is the process of changing one form of energy to another form...

Irina18 [472]3 years ago

4 0

Energy transformation or energy conversion is the process of changing one form of energy to another form ofenergy. In physics, the term energydescribes the capacity to produce certain changes within any system, without regard to limitations intransformation imposed

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What happens to the gravitational potential energy between two particles if the distance between them is halved? (a) It does not

mr_godi [17]

Answer:

The gravitational potential energy between two particles, if the distance between them is halved, is multiplied by 4 (option c).

Explanation:

The gravitational force is the force of mutual attraction that two objects with mass experience.

The Law of Universal Gravitation enunciated by Newton says that every material particle attracts any other material particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance that separates them. Mathematically this is expressed as:

$F=G*\frac{m1*m2}{r^{2} }$

where m1 and m2 are the masses of the objects, r the distance between them and G a universal constant that receives the name of constant of gravitation.

If the distance between two particles is reduced by half, then, where F' is the new value of the gravitational force:

$F'=G*\frac{m1*m2}{(\frac{r}{2} )^{2} }$

$F'=G*\frac{m1*m2}{\frac{(r )^{2} }{2^{2} } }$

$F'=G*\frac{m1*m2}{\frac{(r )^{2} }{4} }$

$F'=4*G*\frac{m1*m2}{r^{2} }$

F'=4*F

The gravitational potential energy between two particles, if the distance between them is halved, is multiplied by 4 (option c).

7 0

3 years ago

A student pushes on a 8-kg box with a force of 35 N forward. The force of sliding friction is 10 N backward. What is the acceler

Ne4ueva [31]

Answer:

(35 N - 10 N)/8kg = 3.125 m/s^2

Explanation:

The formula for Force is:

Force = Mass*Acceleration

(Force is equal to Mass times Acceleration)

Since we're told to find the acceleration of the box. We make acceleration the subject of the equation:

Acceleration = Force/Mass

(Acceleration equal to Force divided by Mass)

We know that the force are 35 N forward and 10 N backward, and the weight of the box is 8kg.

= (35 N - 10 N)/8kg

The reason that 35 N minus 10 N is because the 10 N is pushing the box backward.

= 25 N/8kg

= 3.125 m/s^2

Hope it helps :DD

3 0

3 years ago

Sphere 1 with radius R_1 has positive charge q, Sphere 2 with radius 4.50 R_1 is far from sphere 1 and initially uncharged. The

tia_tia [17]

Answer:

Explanation:

capacitance of sphere 2 will be 4.5 times sphere 1

a ) when spheres are in contact they will have same potential finally . So

V_1 / V_2 = 1

b )

Charge will be distributed in the ratio of their capacity

charge on sphere1 = q x 1 / ( 1 + 4.5 )

= q / 5.5

fraction = 1 / 5.5

c ) charge on sphere 2

= q x 4.5 / 5.5

fraction = 4.5 / 5.5

d ) surface charge density of sphere 1

= q /( 5.5 x A ) where A is surface area

surface charge density of sphere 2

= q x 4.5 /( 5.5 x 4.5² A ) where A is surface area

= q /( 5.5 x 4.5 A )

q_1/q_2 = 4.5

6 0

2 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

2-D Projectile Motion

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$