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4 years ago

E=MC2

Physics

2 answers:

Bond [772]4 years ago

6 0

These items would BEST fit in section Bof this Venn Diagram. These are listed characteristics of both fission and fusion.

nasty-shy [4]4 years ago

3 0

A. Fission. Remember that most of the energy inside an atom is concentrated on the nucleus, and also most of its mass. There are actually forces that binds the subatomic particles of protons and neutrons inside the nucleus, and this forces are extremely powerful. Protons has a positive charge, there are about 90 inside a Plutonium atom,for example, and this protons should be repelling each other, but they are being bind by what physicist call "strong nuclear force". During nuclear fission a neutron atom will penetrate the nucleus of plutonium atom and release tremendous amount of energy. Multiply the mass of any material to the square of the speed of light. A 10 kg mass multiply by the square of the speed of light, will give you an enormous amount of energy in unit of Joule. E=MC2 That is matter energy equivalence principle. Energy is matter and matter is energy. This equation also prevent us from travelling at the speed of light.

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An ant walks 2\,\text m2m2, start text, m, end text to the right and then climbs 1 \,\text m1m1, start text, m, end text up a wa

NNADVOKAT [17]

Answer:

26.56°

Explanation:

given,

horizontal distance travel by the ant = 2 m

vertical distance travel by the ant = 1 m

angle of the ant above the initial point = ?

using trigonometric function

$tan \theta = \dfrac{h}{x}$

θ is the angle of the ant form initial position

h is the height on the wall

x is the horizontal distance covered by the ant

$tan \theta = \dfrac{1}{2}$

$\theta =tan^{-1}(0.5)$

$\theta =26.56^0$

hence, the angle made by the ant from the initial position is equal to 26.56°

3 0

3 years ago

HELP!!

QveST [7]

Answer:

Total displacement 155 m

Explanation:

From the question, the formula to apply will be that of cosine rule where:

$c=\sqrt{a^2+b^2-2ab*cosC}$

where

a= 43 m , b=130m and c =? , <C=28° + 90° =118°

Substituting values in the equation as;

$c=\sqrt{43^2+130^2-2*43*130*cos118}$

$c=\sqrt{1849+16900-11180*-0.46947} \\\\c=\sqrt{1849+16900+5249} \\\\c=\sqrt{23998} \\\\\\c=155m$

6 0

3 years ago

Which term describes the image of an object that is placed in front o a convex mirror?

Dmitriy789 [7]

Answer:

The answer of this question is :- Virtual image

5 0

3 years ago

Read 2 more answers

when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is remo

Lera25 [3.4K]

Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

$F = k\frac{|q_1|.|q_2|}{r^2}$

Where,

k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

$\frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1$ ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

$q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}$

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

$\frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2} = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 = \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6$ .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

$-\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123$

$q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\$

6 0

4 years ago

Determine the velocity (in m/s) of the object during the last six seconds. Include a numerical answer accurate to the second dec

Sveta_85 [38]

Answer:

0.33 m/s

Explanation:

The graph given is that of distance against time. The slope of such a graph gives the velocity of the object while travelling.

In this graph the change in distance for the last 6 seconds is given by;

The distance is : 10-8 = 2 m

The time in seconds is: 6 s

The velocity = 2/6 = 1/3 m/s

The velocity expressed in decimal form is : 0.33 m/s

6 0

3 years ago