Physics College

Which of these is NOT an effect of humor?


strengthened immune system

reduced stress levels

reduced feelings of anxiety

feelings of jealousy and envy

Answers

Answer 1
feelings of jealousy and envy

Related Questions

Consider an electron confined in a region of nuclear dimensions (about 5 fm). Find its minimumpossible kinetic energy in MeV. Treat this problem as one-dimensional, and use the relativistic relationbetweenEandp. Give your answer to 2 significant figures. (The large value you will find is a strongargument against the presence of electrons inside nuclei, since no known mechanism could contain anelectron with this much energy.)

Answers

Answer:

39.40 MeV

Explanation:

Determine the minimum possible Kinetic energy

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

A metallic circular plate with radius r is fixed to a tabletop. An identical circular plate supported from above by a cable is fixed in place a distance d above the first plate. Assume that d is much smaller than r. The two plates are attached by wires to a battery that supplies voltage V.

Required:
a. What is the tension in the cable?
b. Compute the energy stored in the electric field after the top plate was raised.

Answers

Answer:

A) F = V²E_o•πr²/2d²

B) U = E_o•Aπr²V²/2d

Explanation:

A) Since we have two circular plates, the formula for the electric field is expressed as;

E = V/d

Where;

V is voltage

d is distance

However, the net electric field produced is given by;

E' = V/2d

The tension in the cable can then be expressed as;

F = qE'

Where q is charge

Thus;

F = qV/2d - - - (eq 1)

We also know that;

C = q/V = E_o•A/d

A is area = πr²

Thus;

q/V = E_o•πr²/d

q = VE_o•πr²/d

Let's put VE_o•πr²/d for q in eq 1 to get;

F = V²E_o•πr²/2d²

B) formula for the energy stored in the electric field is;

U = ½CV²

From earlier, we saw that; C = E_o•A/d

Thus;

U = ½E_o•AV²/d

A = πr²

Thus;

U = E_o•Aπr²V²/2d

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