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Komok [63]
3 years ago
9

Define A Gamma Ray .....---_---__☘️​

Physics
1 answer:
MrRa [10]3 years ago
6 0

Answer:

gamma ray, or gamma radiation (symbol γ or {\displaystyle \gamma } \gamma ), is a penetrating form of electromagnetic radiation arising from the radioactive decay of atomic nuclei. It consists of the shortest wavelength electromagnetic waves and so imparts the highest photon energy. Paul Villard, a French chemist and physicist, discovered gamma radiation in 1900 while studying radiation emitted by radium. In 1903, Ernest Rutherford named this radiation gamma rays based on their relatively strong penetration of matter; in 1900 he had already named two less penetrating types of decay radiation (discovered by Henri Becquerel) alpha rays and beta rays in ascending order of penetrating power.

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D multiply force

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Cuando una persona sube y baja una escalera, Cuanto vale su desplazamiento y cual es la medida de su trayectoria.
adoni [48]

Answer:

Primero, definimos el desplazamiento como la distancia entre la posición final y la posición inicial.

Así, si comenzamos abajo, luego subimos la escalera, y luego bajamos, la posición final y la posición inicial serán la misma

por lo que el desplazamiento es igual a cero.

La medida recorrida es el espacio total recorrido.

Es decir, si entre el principio y el final de la escalera hay una distancia D.

La persona que sube y baja, recorre esta distancia dos veces.

Entonces cuando una persona sube y baja la escalera, la medida de su trayectoria será 2*D.

8 0
3 years ago
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Sonbull [250]

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1.Stronger bones 2.Joint flexibility

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2 years ago
What would happen if our bodies could not metabolize glucose?
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3 years ago
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A thermos contains m1 = 0.89 kg of tea at T1 = 31° C. Ice (m2 = 0.075 kg, T2 = 0° C) is added to it. The heat capacity of both w
lorasvet [3.4K]

Answer:

a) T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

b) T = 295.37 K

Explanation:

Given;

Initial temperature of tea T1 = 31 C

Initial temperature of ice T2 = 0 C

Mass of tea m1 = 0.89 kg

Mass of ice m2 = 0.075kg

The heat capacity of both water and tea c = 4186 J/(kg⋅K)

the latent heat of fusion for water is Lf = 33.5 × 10^4 J/kg

And T = the final temperature of the mixture

Heat loss by tea = heat gained by ice

m1c∆T1 = m2c∆T2 + m2Lf

m1c(T1-T) = m2c(T-T2) + m2Lf

m1cT1 - m1cT = m2cT - m2cT2 + m2Lf

m1cT + m2cT = m1cT1 + m2cT2 - m2Lf

T(m1c + m2c) = m1cT1 + m2cT2 - m2Lf

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

Substituting the values;

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

T = (0.89×4186×31 + 0.075×4186×0 - 0.075×33.5 × 10^4)/(0.89×4186 + 0.075×4186)

T = 22.37 °C

T = 273 + 22.37 K

T = 295.37 K

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3 years ago
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