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2 years ago

PLEASE HELP! ASAP

Physics

2 answers:

hoa [83]2 years ago

8 0

Explanation:

Specifically, small amounts of mass are turned into energy from the breaking up (fission) or combination (fusion) of the nuclei of atoms. Even spontaneous radioactive decay converts a bit of mass into incredible amounts of energy.

Damm [24]2 years ago

5 0

Answer:

A. Uranium breaks apart into krypton, barium, and three neutrons.

Explanation:

did the test !!

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Which element is stable and nonreactive?

const2013 [10]

The answer is D. Krypton (kr)

6 0

3 years ago

18. Un avión de rescate de animales que vuela hacia el este a 36.0 m/s deja caer una paca de

Alona [7]

Answer:

Definimos momento como el producto entre la masa y la velocidad

P = m*v

(tener en cuenta que la velocidad es un vector, por lo que el momento también será un vector)

Sabemos que el peso de la paca de heno es 175N, y el peso es masa por aceleración gravitatoria, entonces.

Peso = m*9.8m/s^2 = 175N

m = (175N)/(9.8m/s^2) = 17.9 kg

Ahora debemos calcular la velocidad de la paca justo antes de tocar el suelo.

Sabemos que la velocidad horizontal será la misma que tenía el avión, que es:

Vx = 36m/s

Mientras que para la velocidad vertical, usamos la conservación de la energía:

E = U + K

Apenas se suelta la caja, esta tiene velocidad cero, entonces su energía cinética será cero y la caja solo tendrá energía potencial (Si bien la caja tiene velocidad horizontal en este punto, por la superposición lineal podemos separar el problema en un caso horizontal y en un caso vertical, y en el caso vertical no hay velocidad inicial)

Entonces al principio solo hay energía potencial:

U = m*g*h

donde:

m = masa

g = aceleración gravitatoria

h = altura

Sabemos que la altura inicial es 60m, entonces la energía potencial es:

U = 175N*60m = 10,500 N

Cuando la paca esta próxima a golpear el suelo, la altura h tiende a cero, por lo que la energía potencial se hace cero, y en este punto solo tendremos energía cinética, entonces:

10,500N = (m/2)*v^2

De acá podemos despejar la velocidad vertical justo antes de golpear el suelo.

√(10,500N*(2/ 17.9 kg)) = 34.25 m/s

La velocidad vertical es 34.25 m/s

Entonces el vector velocidad se podrá escribir como:

V = (36 m/s, -34.25 m/s)

Donde el signo menos en la velocidad vertical es porque la velocidad vertical es hacia abajo.

Reemplazando esto en la ecuación del momento obtenemos:

P = 17.9kg*(36 m/s, -34.25 m/s)

P = (644.4 N, -613.075 N)

6 0

3 years ago

A gas in a closed container is heated with 12J of energy, causing the lid of the container to rise 3m with 5N of force. What is

levacccp [35]

Answer:

27J

Explanation:

From conservation of Thermal energy, the total internal energy is the total sum of energy supplied or taken from the system plus work done for or on the system.

Now the change in internal energy would be the sum of the received energy substended in the gas plus the work done by the system which is workdone that it will sustend in pushing the lid. This is expressed mathematically as;

U = Q + (F×d);

U- change in internal energy

Q is the energy received by the system and is positive when energy is received by the system.

Fxd is the workdone and is positive since the gas pushes up the lid- the system does work.

U=12+(3×5)= 27J

5 0

3 years ago

Complete the equation to show the radioactive decay of carbon-14 to nitrogen-14

blsea [12.9K]

Answer:

The beta decay takes place.

Explanation:

The reaction of radioactivity of carbon 14 to nitrogen 14 is

There is a beta decay.

The reaction is

$C_{6}^{14}\rightarrow N_{7}^{14}+\beta _{-1}^{0}+ energy$

Here some energy is released in form of neutrino.

7 0

2 years ago

In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.

vaieri [72.5K]

Answer:

(A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

Explanation:

Given that,

A nucleus undergoes a nuclear decay.

(A). In alpha decay,

We know that,

When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.

We need to calculate the resulting nucleus

Using given data

$^{227}_{89}Ac\Rightarrow ^{227-4}_{89-2}X$

$^{227}_{89}Ac\Rightarrow ^{223}_{87}Fr$

The resulting nucleus is Fr.

(B). In beta-minus decay,

We know that,

When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.

We need to calculate the resulting nucleus

Using given data

$^{211}_{83}Bi\Rightarrow ^{211}_{83+1}X$

$^{211}_{83}Bi\Rightarrow ^{211}_{84}Po$

The resulting nucleus is Po.

(C). In beta-plus decay,

We know that,

When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.

We need to calculate the resulting nucleus

Using given data

$^{22}_{11}Na\Rightarrow ^{22}_{11-1}X$

$^{22}_{11}Na\Rightarrow ^{22}_{10}Ne$

The resulting nucleus is Ne.

(D). In gamma decay,

We know that,

When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.

We need to calculate the resulting nucleus

Using given data

$^{98}_{43}Tc\Rightarrow ^{98}_{43}Tc$

The resulting nucleus is Tc.

Hence, (A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

4 0

3 years ago