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3 years ago

When nuclear fission occurs some mass is lost where does that mass go

Physics

2 answers:

Marianna [84]3 years ago

7 0

Sample Response: Some of the mass is converted to energy. The relative amounts of matter and energy stay the same.

horrorfan [7]3 years ago

4 0

The loss of matter is called the mass defect. The missing matter is converted into energy. You can actually calculate the amount of energy produced during a nuclear reaction with fairly simple equation developed by Albert Einstein; E = mc^2. In this equation, E is the amount of energy produced, m is the missing mass, or the mass defect, and c is the speed of light, which is a rather large number. The speed of light is squared, making that part of the equation a very large number that, even when multiplied by a small amount of mass, yields a large amount of energy.

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svetlana [45]

Answer:

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2 years ago

What is the full meaning of (i.p.s.m.n)

Anna11 [10]

Answer:

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3 years ago

Which object had more potential energy when it was lifted to a distance of 1000 centimeters? Show your calculation.

RSB [31]

Explanation:

We Know That

POTENTIAL ENERGY= MASS*g*HEIGHT

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2 years ago

If the car ha a mass of 1000 kilograms, what is its momentum (v=35m/s)

Black_prince [1.1K]

Answer:

The momentum of the car is 35,000 kg.m/s

Explanation:

Momentum

Momentum is often defined as mass in motion.

Since all objects have mass, if it's moving, then it has momentum. It can be calculated as the product of the mass by the velocity of the object:

$\vec p = m\vec v$

If only magnitudes are considered:

p = mv

The car has a mass of m=1,000 kg and travels at v=35 m/s. Calculating its momentum:

p = 1,000 kg * 35 m/s

p = 35,000 kg.m/s

The momentum of the car is 35,000 kg.m/s

4 0

3 years ago

Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)

sergeinik [125]

$v^2-{v_0}^2=2a(x-x_0)$

dónde $v$ es la velocidad de la esfera, $v_0$ es suya velocidad inicial, $a=-g$ la aceleración debida a la gravedad, $x$ la posición, y $x_0$ la posición inicial. Tomamos $x_0=0\,\mathrm m$ a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

$v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)$

$\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}$

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

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3 years ago