Answer:
Explanation:
It is a case of adiabatic expansion .
[tex]T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}[/tex]
T₁ , T₂ are initial and final temperature , V₁ and V₂ are initial and final volume.
Given ,
V₂ = 3 V₁ and T₁ = 292 . γ for air is 1.4 .
[tex]( 3 )^{\gamma-1}= \frac{292}{ T_2}[/tex]
[tex]( 3 )^{1.4-1}= \frac{292}{ T_2}[/tex]
1.552 = 292 / T₂
T₂ = 188 K .
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.)
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.
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
Who is a socio-economically disadvantaged child? Explain any four ways forhelping such a child
Explanation:
A socio-economically disadvantaged child is the one who is disadvantaged in terms of social position and economic position. Such children have limited resources in terms of education, money and future options. Four ways of helping such children are as follows:
1) Help them in education - You can help such children by giving them free tuition. If you belong to a well off family, you can get them admitted in schools as well. Provide them with books and uniform.
2) Encourage them to do well in school and pursue their passions.
3) Provide them with meals, if they do not have access to regular meals.
4) encourage them to go school regularly.
Answer:
sorry i dont know the answer bit mark me as BRAINLIST