-1
If 5*(p + 6) = 25 then p is -1
Suppose the perfect number P has divisors f1, f2, ... , fk, P (where all the fs are smaller than P. Then, by the definition of a perfect number, f1 + f2 + ... + fk = P So that f1 + f2 + ... + fk + P = 2P Dividing through by P, f1/P + f2/P + ... + fk/P + P/P = 2 . . . . . . . . . . . (2) Now, since f1 is a factor of P, the f1*g1 = P where g is the cofactor and so f1/P = 1/g1. That is, f1/P is the reciprocal of one of the other factors of P. Also, the last term on the left is the reciprocal of the factor 1. And therefore, f1/P + f2/P + ... + fk/P + P/P = 2 = 1/g1 + 1/g2 + ... + 1/gk + 1/1
5 p/1 pound = 5 p/100 p = 1/20
No, it is not.Incidentally, is a number p is prime then its 2 factors are 1 and p. Their sum is p+1.If p is a perfect number then this sum must be 2p.That is, p+1 = 2p or p = 1.But 1 is not a prime. So no prime can be a perfect number and since 13 is prime, it cannot be a perfect number.
It has one nucleon. It has no neutrons. It has one proton.
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Proton.
I am assuming that this is to do with the trajectory that is simplified to that of a particle which does not incur air resistance. If I have understood the question correctly, the particle travels under the influence of a constant force - assumed to be gravity which acts downwards. The model can be extended to allow for a constant force acting at an angle but the calculations then become more complicated. The particle is projected upwards, with the initial velocity, u ms-1, which makes an angle P with the horizontal. u is a variable such that the horizontal range of the particle is a constant. The vertical component of the initial velocity is u*sin(P) ms-1. The gravitational force, acting downwards, is -g ms-2. When the particle returns to the ground level, the vertical component of its velocity is -u*sin(P) ms-1. So if the particle returns at time t seconds, then t = [u*sin(P) - -u*sin(P)] /g = 2*u*sin(P)/g sec. The horizontal component of the velocity of the particle is a constant u*cos(P) ms-1. So during the time in flight, it travels u*cos(P)*2*u*sin(P)/g m = 2*u2*sin(P)*cos(P)/g m. This horizontal distance is constant, which implies that 2*u2*sin(P)*cos(P)/g is constant so that u2 is inversely proportional to sin(P)*cos(P). So let u = sqrt[k/sin(P)*cos(P)] ms-1 for some constant k. then its vertical component is u*sin(P) = sqrt[k/sin(P)*cos(P)]*sin(P) ms-1 = sqrt[k*tan(P)] Then at time T, its height is sqrt[k*tan(P)]*T - 0.5g*T2 I just hope this is correct!
The answer is 1/90.
When P-32 decays to S-32, a beta particle is emitted. This beta particle is an electron released during the conversion of a neutron into a proton within the nucleus of the atom.
Particle
A gamma ray is not an elementary particle, it is a high-energy photon. Photons are elementary particles that have zero rest mass and carry electromagnetic radiation.
particle
Particle velocityparticle displacementpolarizationpressureplasmas
Particle.