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Aleksandr [31]
3 years ago
9

A 0.22-caliber handgun fires a 28-g bullet at a velocity of 765 m/s. Calculate the de Broglie wavelength of the bullet.

Physics
1 answer:
ser-zykov [4K]3 years ago
8 0

Answer:

de Broglie wavelength of the bullet is 3.093 x 10⁻³⁵ m

Explanation:

Given;

mass of bullet, m = 28 g = 0.028 kg

velocity of the bullet, v = 765 m/s

de Broglie wavelength of the bullet is given by;

λ = h / mv

where;

λ is de Broglie wavelength of the bullet

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

λ = h / mv

λ = (6.626 x 10⁻³⁴ ) / (0.028 x 765)

λ = 3.093 x 10⁻³⁵ m

Therefore, de Broglie wavelength of the bullet is 3.093 x 10⁻³⁵ m

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A rock is attached to the left end of a uniform meter stick that has the same mass as the rock. How far from the left end of the
kotykmax [81]

Answer:

M₂ = M  then L₂ = L

M₂> M  then L₂ = \frac{M}{M_{2}} L

Explanation:

This is a static equilibrium exercise, to solve it we must fix a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive

          ∑ τ = 0

           

The mass of the rock is M and placed at a distance, L the mass of the rod M₁, is considered to be placed in its center of mass, which by uniform e is in its geometric center (x = 0) and the triangular mass M₂, with a distance L₂

The triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.

         M L + M₁ 0 - m₂ L₂ = 0

         M L - m₂ L₂ = 0

         L₂ = \frac{M}{M_{2}} L

From this answer we have several possibilities

* if the two masses are equal then L₂ = L

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6 0
3 years ago
Which of the following is not a natural resource? a. time b. water c. land d. air Please select the best answer from the choices
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Answer: Option (a) is the correct answer.

Explanation:

Resources which are created by human beings are known as man-made resources.

For example, glass, rayon, nylon etc are all man-made resources.

Whereas resources which are naturally created are known as natural resources.

For example, wind, air, water etc are all natural resources.

Thus, we can conclude that out of the given options time is not a natural resource.

3 0
3 years ago
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar
defon

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

The power output W = 2 MW

Efficiency \zeta = 3%

where;

\zeta = \dfrac{W_{out}}{Q_{supplied }}

Q_{supplied } = \dfrac{2}{0.03} \ MW

Q_{supplied } = 66.66 \ MW

However, from the evaporator, the heat transfer Q can be determined by using the formula:

Q = UA(L MTD)

where;

LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}

Also;

\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K

\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K

LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}

LMTD = \dfrac{8}{In (5)}

LMTD = 4.97

Thus, the required heat exchanger area A is calculated by using the formula:

Q_H = UA (LMTD)

where;

U = overall heat coefficient given as 1200 W/m².K

66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\  A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\  \mathbf{A = 11178.236 \ m^2}

The mass flow rate:

Q_{H} = mC_p(T_{in} -T_{out} )  \\ \\  66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{  66.667 \times 10^6}{4.18 \times 8} \\ \\  \mathbf{m = 1993630.383 \ kg/s}

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