The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for <span>52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=</span><span>binding energy
</span><span>52He-----------> 2 x 11p + 3 x10n is the equation bilan
</span>so Dm=2 mp + (5-2)mn-mnucleus, mp=mass of proton=1.67 10^-27 kg
mn=mass of neutron=<span>1.67 10^-27 kg
</span><span>m nucleus= 5
Dm= 2x</span>1.67 10^-27 kg+ 3x<span>1.67 10^-27 kg-5= - 4.9 J
Eb= </span> - <span>4.9 J x c²= -4.9 x 9 .10^16= - 45 10^16 J
so the answer is Eb /nucleon = Eb/5= -9.10^16 J, but 1eV=1.6 . 10^-19 J
so </span><span>-9.10^16 J/ 1.6 10^-19= -5.625 10^35 eV
the final answer is </span><span>Eb /nucleon </span><span>= -5.625 x10^35 eV</span>
A.) reference group
"A reference group includes individuals or groups that influence our opinions, beliefs, attitudes and behaviors. They often serve as our role models and inspiration"(study.com).
Responder:
20πrads ^ -1; 24πrads ^ -1; 0,1 seg; 10 Hz
Explicación:
Dado lo siguiente:
Radio (r) del círculo = 120 cm
600 revoluciones por minuto en radianes por segundo
(600 / min) * (2π rad / 1 rev) * (1min / 60seg)
(1200πrad / 60sec) = 20π rad ^ -1
Velocidad angular (w) = 20πrads ^ -1
Velocidad lineal = radio (r) * velocidad angular (w)
Velocidad lineal = (120/100) * 20πrad
Velocidad lineal = 1.2 * 20πrads ^ -1 = 24πrads ^ -1
C.) Período (T):
T = 2π / w = 2π / 20π = 0.1 seg
D.) Frecuencia (f):
f = 1 / T = 1 / 0.1
1 / 0,1 = 10 Hz
Answer:71 dB
Explanation:
Given
sound Level 
distance 
From sound Intensity
we know Intensity 
Sound level corresponding to 
Between magnitude of the average 4sec
