Why are there no hurricanes in Alaska?

The trough of the sine curve used to represent a sound wave corresponds to

iren [92.7K]

Answer:

The correct answer is a rarefaction.

Explanation:

Sound waves are longitudinal waves that propagate in a medium, such as air. As the vibration continues, a series of successive condensations and rarefactions form and propagate from it. The pattern created in the air is something like a sinusoidal curve to represent a sound wave.

There are peaks in the sine wave at the points where the sound wave has condensations and valleys where it has rarefactions.

Have a nice day!

4 0

4 years ago

In a particular experiment to study the photoelectric effect, the frequency of the incident light and the temperature of the met

qwelly [4]

Answer:

B. The number of electrons emitted from the metal per second increases.

Explanation:

Light consists of photons . Energy of each photon depends upon frequency of light . The increase in intensity increases the number of photons . It does not increase energy of photons .

So if a high intensity light falls on a photosensitive plate , each photon ejects one electron . So number of electrons increases if we increase intensity of photon. It does not increase kinetic energy of ejected electrons . Work function depends upon the nature of plate.

6 0

3 years ago

What is the total number of atoms in the following formula? CaBr2 4 3 2 1

ololo11 [35]

Your answer is 3 ( 1 calcium atom and 2 bromine atoms)

8 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:

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

3 0

3 years ago

A projectile is fired from ground level with an initial speed of 55.6 m/s at an angle of 41.2° above the horizontal. (a) Determi

sattari [20]

Answer:

(a) t = 3.74 s

(b) H = 136.86 m

(c) Vₓ = 41.83 m/s, Vy = 0 m/s

(d) ax = 0 m/s², ay = 9.8 m/s²

Explanation:

(a)

Time to reach maximum height by the projectile is given as:

t = V₀ Sinθ/g

where,

V₀ = Launching Speed = 55.6 m/s

Angle with Horizontal = θ = 41.2°

g = 9.8 m/s²

Therefore,

t = (55.6 m/s)(Sin 41.2°)/(9.8 m/s²)

t = 3.74 s

(b)

Maximum height reached by projectile is:

H = V₀² Sin²θ/g

H = (55.6 m/s)² (Sin²41.2°)/(9.8 m/s²)

H = 136.86 m

(c)

Neglecting the air resistance, the horizontal component of velocity remains constant. This component can be evaluated by the formula:

Vₓ = V₀ₓ = V₀ Cos θ

Vₓ = (55.6 m/s)(Cos 41.2°)

Vₓ = 41.83 m/s

Since, the projectile stops momentarily in vertical direction at the highest point. Therefore, the vertical component of velocity will be zero at the highest point.

Vy = 0 m/s

(d)

Since, the horizontal component of velocity is uniform. Thus there is no acceleration in horizontal direction.

ax = 0 m/s²

The vertical component of acceleration is always equal to the acceleration due to gravity during projectile motion:

ay = 9.8 m/s²

3 0

3 years ago