How much work is done on 10.0C of charge to move it through a potential difference of 9V in 10s?

A pendulum is swinging next to a wall. The distance from the bob of the pendulum to the wall varies in a periodic way that can b

nexus9112 [7]

The formula of the trigonometric function that models the distance HHH from the pendulum's bob to the wall after t seconds is

H(t) = 15 -6sin(2.5π(t -0.5))

Detailed explanation:

The function can be expressed as the following for the midline M, amplitude A, period T, and time t0 at which the function deviates from the midline:

H(t) = M -Asin(2π/T(t -t0))

The equation is based on the parameters M=15, A=6, T=0.8, and t0 = 0.5.

H(t) is equal to 15 -6sin(2.5π(t -0.5))

<h3>What is function?</h3>

The trigonometric functions in mathematics are real functions that connect the right-angled triangle's angle to the ratios of its two side lengths .In all areas of study that involve geometry, such as geodesy, solid mechanics, celestial mechanics, and many others, they are widely used.

To learn more about functions visit:

brainly.com/question/15607563

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The correct question is:

A pendulum is swinging next to a wall. The distance from the bob of the swinging pendulum to the wall varies in a periodic way that can be modeled by a trigonometric function.

The function has period 0.80.80, point, 8 seconds, amplitude 6 \text{ cm}6 cm6, start text, space, c, m, end text, and midline H = 15 \text{ cm}H=15 cmH, equals, 15, start text, space, c, m, end text. At time t = 0.5t=0.5t, equals, 0, point, 5 seconds, the bob is at its midline, moving towards the wall.

Find the formula of the trigonometric function that models the distance HHH from the pendulum's bob to the wall after t seconds. Define the function using radians.

5 0

1 year ago

What is a non-example of Newton's 3rd law of motion?

kirill [66]

Answer:

Unbalanced forces are not examples of Newton's third law because not all opposite reactions are unbalanced forces.

Explanation:

4 0

2 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 surface recieving sound is moved from it original position to a position three times farther away from the source of the sound

Delicious77 [7]

Sound intensity = 1/(r^2)

That is Sound intensity is indirectly proportional to the distance. Therefore, sound becomes 9 times less intense.

7 0

2 years ago

Different ____ of light through two separate mediums causes the bending of waves fronts associated with light rays.

Fittoniya [83]

Different speeds of light through two separate media ... and the difference in wavelength caused by the difference ... causes the bending of waves fronts associated with light rays.

AFTER the bend, since the light rays then travel in a different direction, we may also say that the 'velocity' has changed.

8 0

2 years ago