35m/s is meters per second.
35x35=1225
it will have gone a total of 1225 meters on thirty five seconds
Answer:
I think its oxygen, a conductor is something that allows heat very easily.
Explanation:
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Answer:
The final charges of each sphere are: q_A = 3/8 Q
, q_B = 3/8 Q
, q_C = 3/4 Q
Explanation:
This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.
Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point
q_A = Q / 2
q_B = Q / 2
Now sphere A touches sphere C, ending with half the charge
q_A = ½ (Q / 2) = ¼ Q
q_B = ¼ Q
Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge
q = Q / 4 + Q / 2 = ¾ Q
This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q
q_A = 3/8 Q
q_B = 3/8 Q
The final charges of each sphere are:
q_A = 3/8 Q
q_B = 3/8 Q
q_C = 3/4 Q
Answer:
Explanation:
a ) y = A sin(B) ; here B is the phase of the wave which moves so that it remains constant
ωt - kx = constant
differentiating on both sides
ωdt - kdx =0
ωdt = kdx
dx / dt = ω / k
wave velocity = ω / k
b ) ω = 14.5 rad / s ,
k = 18 rad / m
wave velocity = ω / k
= 14.5 / 18
= .805 m /s
80.5 cm / s
c )
Amplitude = A
= 9.5 m
Answer:
39.40 MeV
Explanation:
<u>Determine the minimum possible Kinetic energy </u>
width of region = 5 fm
From Heisenberg's uncertainty relation below
ΔxΔp ≥ h/2 , where : 2Δx = 5fm , Δpc = hc/2Δx = 39.4 MeV
when we apply this values using the relativistic energy-momentum relation
E^2 = ( mc^2)^2 + ( pc )^2 = 39.4 MeV ( right answer ) because the energy grows quadratically in nonrelativistic approximation,
Also in a nuclear confinement ( E, P >> mc )
while The large value will portray a Non-relativistic limit as calculated below
K = h^2 / 2ma^2 = 1.52 GeV