Answer:
A rotation occurs after every side out, which is when the receiving team gains the right to serve by winning a rally. ... The new serving team will rotate clockwise one spot. The purpose of this is to rotate all the players through the serving position. If you continue winning points, you stay in position.
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>
Answer:
Explanation:
The question is incomplete.
The equation of motion is given for a particle, where s is in meters and t is in seconds. Find the acceleration after 4.5 seconds.
s= sin2(pi)t
Acceleration = d²S/dt²
dS/dt = 2πcos2πt
d²S/dt² = -4π²sin2πt
A(t) = -4π²sin2πt
Next is to find acceleration after 4.5 seconds
A(4.5) = -4π²sin2π(4.5)
A(4.5) = -4π²sin9π
A(4.5) = -4π²sin1620
A(4.5) = -4π²(0)
A(4.5) = 0m/s²
Answer:
Explanation:
extension in the spring = 40.4 - 31.8 = 8.6 cm = 8.6 x 10⁻² m .
kx = mg
k is spring constant , x is extension , m is mass
k x 8.6 x 10⁻² = 7.52 x 9.8
k = 856.93 N/m
= 857 x 10⁻³ KN /m
b ) Both side is pulled by force of 188 N .
Tension in spring = 188N
kx = T
856.93 x = 188
x = .219.38 m
= 21.938 cm
= 21.9 cm .
length of spring = 31.8 + 21.9
= 53.7 cm .
Answer:
Fy=107.2 N
Explanation:
Conceptual analysis
For a right triangle :
sinβ = y/h formula (1)
cosβ = x/h formula (2)
x: side adjacent to the β angle
y: opposite side of the β angle
h: hypotenuse
Known data
h = T = 153.8 N : rope tension
β= 44.2°with the horizontal (x)
Problem development
We apply the formula (1) to calculate Ty : vertical component of the rope force.
sin44.2° = Ty/153.8 N
Ty = (153.8 N ) *(sen44.2°)= 107.2 N directed down
for equilibrium system
Fy= Ty=107.2 N
Fy=107.2 N upward component of the force acting on the stake
