The chemical bonds alow it to pass to the other object.
Answer
given,
D = 50 mm = 0.05 m
d = 10 mm = 0.01 m
Force to compress the spring
F = 3160 N
stress correction factor from stress correction curve is equal to 1.1
now, calculation of corrected stress
= 442.6 Mpa
The tensile strength of the steel material of ASTM A229 is equal to 1300 Mpa
now,
since corrected stress is less than the 
hence, spring will return to its original shape.
The number of protons plus the number of neutrons decreases
Beta decay occurs when, in a nucleus with too many protons or too many neutrons, one of the protons or neutrons is transformed into the other. In beta minus decay, a neutron decays into a proton, an electron, and an antineutrino: n Æ p + e - +.
Answer:
v = 3.27 m/s
Explanation:
KE = 1/2 mv^2
695 J = 1/2 (130kg)(v^2)
695 J / (1/2 x 130kg) = v^2
v^2 = square root of 10.69
v = 3.27 m/s
Answer:
speed of the charge electric is v = - (Eo q/m) cos t
Explanation:
The electric charge has a very small mass so it follows the oscillations of the electric field. We force ourselves on the load,
F = q Eo sint
a) To find the velocity of the particle, let's use Newton's second law to find the acceleration and of this by integration the velocity
F = ma
q Eo sint = ma
a = Eo q / m sint
a = dv / dt
dv = adt
∫ dv = ∫ a dt
v-vo = I (Eoq / m) sin t dt
v- vo = Eo q / m (-cos t)
We evaluate the integral from the initial point, as the particle starts from rest Vo = 0, for t = 0
v = - (Eo q / m) cos t
b) Kinetic energy
K = ½ m v2
K = ½ m (Eoq / m)²2 (sint)²
K = ¹/₂ Eo² q² / m sin² t
c) The average kinetic energy over a period
K = ½ m v2
<v2> = (Eoq / m) 2 <cos2 t>
The average of cos2 t = ½, substitute and calculate
K = ½ m (Eoq / m)² ½
K = ¼ Eo² q² / m
