A car starts from rest and drives at 45 m/s in 20 s. how far did it travel

The ground state energy of an electron in a one-dimensional trap with zero potential energy in the interior and infinite potenti

ElenaW [278]

Answer:

0.5 eV

Explanation:

$E_1$ = Initial potential energy = $2\ eV$

$E_2$ = Final potential energy

$L_1$ = Initial width

$L_2$ = Final width = $2L_1$

Energy of an electron in a one-dimensional trap is given by

$E=\dfrac{n^2h^2}{8mL^2}$

From the equation we get

$E\propto \dfrac{1}{L^2}$

So,

$\dfrac{E_1}{E_2}=\dfrac{L_2^2}{L_1^2}\\\Rightarrow E_2=\dfrac{E_1L_1^2}{L_2^2}\\\Rightarrow E_2=\dfrac{2L_1^2}{4L_1^2}\\\Rightarrow E_2=0.5\ eV$

The ground state energy will be 0.5 eV

6 0

3 years ago

how much thermal energy is absorbed by a copper water pipe with a mass of 2.3kg when its temperature is raised from 20.0 to 80.0

Sphinxa [80]

Explanation:

q = mCΔt

q = (2.3 kg) (385 J/kg/K) (80.0°C − 20.0°C)

q ≈ 53,000 J

q ≈ 53 kJ

3 0

2 years ago

Сс<br> Explain the effect of Earth's gravity on objects on the<br> surface of Earth.

dem82 [27]

Answer:

The effect of gravity on object makes objects stable and not floating in the air . It also makes any object thrown up to return back

6 0

2 years ago

A driver fills an 18.9L steel gasoline can with gasoline at 15.0°C right up to the top. He forgets to replace the cap and leaves

GarryVolchara [31]

Answer:

ΔV = 0.98 L

Explanation:

First, we will calculate the increased volume using Charles' Law:

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

where,

V₁ =initial volume = 18.9 L

V₂ = final volume = ?

T₁ = initial temperature = 15°C + 273 = 288 k

T₂ = final temperature = 30°C + 273 = 303 k

Therefore,

$\frac{18.9\ L}{288\ k} = \frac{V_2}{303\ k}$

V₂ = 19.88 L

Now, we calculate the change in volume:

ΔV = V₂ - V₁ = 19.88L - 18.9 L

This is the volume of gasoline that will spill out.

5 0

2 years ago

A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to

Nostrana [21]

Answer:

a) r=4.24cm d=1 cm

b) $Q=5x10^{-10} C$

Explanation:

The capacitance depends only of the geometry of the capacitor so to design in this case knowing the Voltage and the electric field

$V=1.00x10^{2}v\\E=1.00x10^{4} \frac{N}{C}$

$V=E*d\\d=\frac{V}{E}\\d=\frac{1.0x10^{2}}{1.0x10^{4}}\\d=0.01m$

The distance must be the separation the r distance can be find also using

$C=\frac{Q}{V_{ab}}$

But now don't know the charge these plates can hold yet so

a).

d=0.01m

$C=E_{o}*\frac{A}{d}\\A=\frac{C*d}{E_{o}}$

$A=\frac{5pF*0.01m}{8.85x10^{-12}\frac{F}{m}}\\A=5.69x10^{-3}m^{2}$

$A=\pi *r^{2}\\r=\sqrt{\frac{A}{r}}\\r=\sqrt{\frac{5.64x10^{-3}m^{2} }{\pi } } \\r=42.55x^{-3}m$

b).

$C=\frac{Q}{V_{ab}}$

$Q=C*V\\Q=5x10^{-12} F*1x10^{2}\\Q=5x10^{-10}C$

5 0

3 years ago