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So we know coefficient of f times normal force is friction. So do 100= .25 times x. Now solve for x. You get 400. So 400 is the normal force. And we know normal force equals weight in these types of problems so the answer is 400
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
We can conclude that star A is closer to us than star B.
In fact, the absolute magnitude gives a measure of the brightness of the star, if all the stars are placed at the same distance from Earth. So, it's a measure of the absolute luminosity of the star, indipendently from its distance from us: since the two stars have same absolute magnitude, it means that if they were at same distance from Earth, they would appear with same luminosity. Instead, we see star A brighter than star B, and the only explanation is that star A is closer to Earth than star B (the closer the star A, the brigther it is)
