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3 years ago

A plane progressive

Physics

1 answer:

kiruha [24]3 years ago

4 0

Answer:

Amplitude, A = 5 m

Explanation:

Let a progressive wave is given by equation :

$y=5\sin (100\pi t-0.4\pi x)$ .....(1)

The general equation of a progressive wave is given by :

$y=A\sin (\omega t-kx)$ ....(2)

Here,

A is the amplitude of the wave

$\omega$ is the angular frequency

k is propagation constant

We need to find the amplitude of the wave.

If we compare equations (1) and (2), we find that the amplitude of the given plane progressive wave is 5 m.

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A light has wavelength of is used in the double slits experiment . Which of the following values represent the phase difference

Wewaii [24]

Answer:

Explanation:

8 0

3 years ago

It took a crew 9 h 36 min to row 8 km upstream and back again. If the rate of flow of the stream was 2 km/h, what was the rowing

babunello [35]

Answer:

3 km/h

Explanation:

Let's call the rowing speed in still water x, in km/h.

Rowing speed in upstream is: x - 2 km/h

Rowing speed in downstream is: x + 2 km/h

It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:

8/(x - 2) + 8/(x + 2) = 48/5 (notice that: time = distance/speed)

Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)

8*(x+2) + 8*(x-2) = (48/5)*(x² - 4)

Dividing by 8

(x+2) + (x-2) = (6/5)*(x² - 4)

2*x = (6/5)*x² - 24/5

0 = (6/5)*x² - 2*x - 24/5

Using quadratic formula

$x = \frac{2 \pm \sqrt{(-2)^2 - 4(6/5)(-24/5)}}{2(6/5)}$

$x = \frac{2 \pm 5.2}{2.4}$

$x_1 = \frac{2 + 5.2}{2.4}$

$x_1 = 3$

$x_2 = \frac{2 - 5.2}{2.4}$

$x_2 = -1\; 1/3$

A negative result has no sense, therefore the rowing speed in still water was 3 km/h

7 0

4 years ago

As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an ax

AysviL [449]

Answer:

$I=2.766\ kg.m^2$

Explanation:

We have:

diameter of the wheel, $d=0.88\ m$

weight of the wheel, $w_w=280\ N$

mass of hanging object to the wheel, $m_o=6.32\ kg$

speed of the hanging mass after the descend, $v_o=4\ m.s^{-1}$

height of descend, $h=2.5\ m$

(a)

moment of inertia of wheel about its central axis:

$I=\frac{1}{2} m.r^2$

$I=\frac{1}{2} \frac{w_w}{g}.r^2$

$I=\frac{1}{2} \times \frac{280}{9.8}\times 0.44^2$

$I=2.766\ kg.m^2$

3 0

4 years ago

An 84.0 kg sprinter starts a race with an acceleration of 1.76 m/s2. If the sprinter accelerates at that rate for 11 m, and then

gulaghasi [49]

Answer:

t=17.838s

Explanation:

The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:

The acceleration is, by definition:

$a=\frac{dv}{dt}=1.76$

So, the velocity can be obtained by integrating this expression:

$v=1.76t$

The velocity is, by definition: $v=\frac{dx}{dt}$ , so

$dx=1.76tdt\\x=1.76\frac{t^{2}}{2}$ .

Do x=11 in order to find the time spent.

$11=1.76\frac{t^2}{2}\\ t^2=\frac{2*11}{76} \\t=\sqrt{12.5}=3.5355s$

At this time the velocity is: $v=1.76t=1.76*3.5355s=6.2225\frac{m}{s}$

This velocity remains constant in the section 2, so for that section the movement equation is:

$x=v*t\\t=\frac{x}{v}$

The left distance is 89 meters, and the velocity is $6.2225\frac{m}{s}$ , so:

$t=\frac{89}{6.2225}=14.303s$

So, the total time is 14.303+3.5355s=17.838s

7 0

3 years ago

A student attempted to measure the specific latent heat of vaporisation of water.

tensa zangetsu [6.8K]

Answer:

The latent heat of vaporization of water is 2.4 kJ/g

Explanation:

The given readings are;

The first (mass) balance reading (of the water) in grams, m₁ = 581 g

The second (mass) balance reading (of the water) in grams, m₂ = 526 g

The first joulemeter reading in kilojoules (kJ), Q₁ = 195 kJ

The second joulemeter reading in kilojoules (kJ), Q₂ = 327 kJ

The latent heat of vaporization = The heat required to evaporate a given mass water at constant temperature

Based on the measurements, we have;

The latent heat of vaporization = ΔQ/Δm

∴ The latent heat of vaporization of water = (327 kJ - 195 kJ)/(581 g - 526 g) = 2.4 kJ/g

The latent heat of vaporization of water = 2.4 kJ/g

6 0

3 years ago