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5 days ago

Let’s consider tunneling of an electron outside of a potential well. The formula for the transmission coefficient is T \simeq e^

{-2CL}T≃e −2CL , where L is the width of the barrier and C is a term that includes the particle energy and barrier height. If the tunneling coefficient is found to be T = 0.050T=0.050 for a given value of LL, for what new value of L\text{'}L’ is the tunneling coefficient T\text{'} = 0.025T’=0.025 ? (All other parameters remain unchanged.) Express L\text{'}L’ in terms of the original LL.

Physics

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For anti-ballistic missile system, the time of flight tf is determined by the initial speed v0 of the missile and the maximum ra

Ostrovityanka [3204]

Refer to the diagram below.

This discussion operates under a basic analysis that overlooks air resistance and variations in the terrain the missile traverses.

Let V₀ be the launch velocity, at an angle θ to the horizontal.
The horizontal velocity component equals V₀ cosθ.
If the flight duration is $t_{f}$ , then
$r=V_{o} \, t_{f}$
where r represents the missile's range.

The time t at which the missile is at ground level is expressed by
$0=V_{o} sin\theta \, t- \frac{1}{2}gt^{2}$
where g signifies acceleration due to gravity.

t = 0 signifies the missile's launch. Thus
$t_{f} = \frac{2V_{o}sin\theta}{g}$

Consequently,
$r= \frac{2V_{o}^{2} sin\theta cos\theta}{g} = \frac{V_{o}^{2} sin(2\theta)}{g}$

Typically, an angle of θ=45° is optimal for achieving maximum range, resulting in
$r= \frac{V_{o}^{2}}{g}$

This discussion applies more accurately to a scud missile than to a powered, guided missile.

Response:
$t_{f} = \frac{r}{V_{o} cos\theta} \\\\ r= \frac{V_{o}^{2} sin(2\theta)}{g}$
Usually, θ=45°

6 0

3 months ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

kicyunya [3294]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

The law of conservation of mass states that the rate of fluid mass ( $m_{1}$ ) entering a system equals the rate at which the fluid mass ( $m_{2}$ ) exits the system.

The mass flow rate can be expressed as follows:

$m = \rho A v$

where $\rho$ denotes the fluid density, $A$ signifies the cross-sectional area through which fluid flows, and $v$ represents the fluid's velocity.

Based on the problem conditions, as the fluid's density remains constant, we can write:

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas for the fluid flow, while $v_{1}$ and $v_{2}$ are the corresponding velocities across those areas.

Given the conditions in the problem, $A_{2} > A_{1}$ , leading from the formula to $v_{2} < v_{1}$ .

Furthermore, fluid pressure arises from the fluid's movement through any specific area. When the fluid accelerates, part of its energy increases its speed in the direction of flow, resulting in lower pressure.

Thus, in this instance, $v_{2} < v_{1}$ the pressure in the larger cross-sectional area $P_{2}$ will exceed the pressure $P_{1}$ in the smaller cross-sectional area, implying

$P_{2} > P_{1}$ .

6 0

3 months ago

Approximating Venus's atmosphere as a layer of gas 50 km thick, with uniform density 21 kg/m3, calculate the total mass of the a

Maru [3345]

Response:

m = 4.9 x 10⁸ kg

Explanation:

The density equation is

ρ = m / V

m = ρ V

the atmosphere's volume equals the volume of the outer atmospheric sphere minus the planetary volume

V = V_atmosphere - V_planet

V = 4/3 π R_atmosphere³ - 4/3 π R_venus³

V = 4/3 π (R_atmosphere³ - R_venus³

)

the planetary radius is R_venus = 6.06 x 10⁶ m.

The radius of the outer atmospheric layer

R_atmosphere = 50 x 10³ + R_venus = 50 x 10³ + 6.06 x 10⁶

R_atmosphere = 6.11 x 10⁶ m

let's calculate the volume

V = 4/3 pi [(6.11 x 10⁶)³ - (6.06 x 10⁶)³]

V = 23.265 x 10⁶ m³

let’s determine the mass

m = 21 23.265 x 10⁶

m = 4.89 x 10⁸ kg

to two significant figures, this is

m = 4.9 x 10⁸ kg

3 0

2 months ago

If 2.0 mol of gas a is mixed with 1.0 mol of gas b to give a total pressure of 1.6 atm, what is the partial pressure of gas a an

serg [3582]

<span>The partial pressure of A = 1.06 atm and the partial pressure of B = 0.53 atm</span>

8 0

2 months ago

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