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3 years ago

A. Label the following properties on the wave below: rarefaction, wavelength, and compression. (Image attached)

Physics

2 answers:

DedPeter [7]3 years ago

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PART a)

Rarefraction : It is the region of lower density where particles move away from each other

Wavelength : distance between two consecutive compression or rarefaction is known as wavelength

Compression : The region of higher density or the region where particles comes closer to each other is compression

Part b)

Here it is given that

$f = 2100 Hz$

$\lambda = 3.0 m$

So here we know that

$wave speed = frequency \times wavelength$

[tex]v = 2100(3) = 6300 m/s

PART c)

As the person drop the stone in the water it creates waves on the surface of water

These waves can move the boat up and down at its own position and it can not displace the boat so it will not move to the shore

PART D)

When stone is dropped in the water it creates waves. these waves when travel through the water and disturb the boat

Due to this disturbance boat starts moving up and down at its own position and it will not displace from its position.

So by this type of waves it will not move towards the shore

Alecsey [184]3 years ago

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a) Rarefactions are in the green frames, compressions in the purple. Wavelength is a distance between same points of the two corresponding rarefactions/compressions.

b) v=wavelength times frequency=3*2100= 6300 m/s.

c) Bad idea that most likely won't work. Waves on water move only in vertical direction. But maybe a high wave will lift the boat and rescue it.

d) Ah, he needs wants the boat reach the shore, I thought it stuck and the person just wants to make it float. Yes, that's not the best idea, see explanation above

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Shkiper50 [21]

The speed that the person needs to leave the ground will be 4.32m/s

From the question given,

Height = 95cm

Since the person leave the ground v = 0m/s

acceleration due to gravity g = 9.8m/s²

Using the equation of motion

v² = u² + 2as

a = -g (upward motion)

s = h (distance changes to height)

The equation will become:

0² = u² - 2gh

0² = u² - 2(9.8)(0.95)

u² = 18.62

u = √18.62

u = 4.32

Hence the speed that the person needs to leave the ground will be 4.32m/s

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Derive the expression for the equivalent resistance of three resistors in[1] paraleli [2] series

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4 years ago

At each corner of a square of side there are point charges of magnitude Q, 2Q, 3Q, and 4Q

Bad White [126]

Answer:

$\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}$

$\displaystyle \theta =68^o$

Explanation:

Electrostatic Force

It's the force that appears between two electrical charges q1 q2 when they are placed at a certain distance d. The force can be computed by using the Coulomb's law:

$\displaystyle F=\frac{KQ_1Q_2}{d^2}$

We have an arrangement of 4 charges as shown in the image below. We need to calculate the total force exerted on the charge 2Q by the other 3 charges. The free body diagram is also shown in the second image provided. The total force on 2Q is the vectorial sum of F1, F2, and F3. All the forces are repulsive, since all the charges have the same sign. Let's compute each force as follows:

$\displaystyle |F_1|=\frac{KQ(2Q)}{l^2}=\frac{2KQ^2}{l^2}$

$\displaystyle |F_2|=\frac{K(2Q)(4Q)}{l^2}=\frac{8KQ^2}{l^2}$

The distance between 3Q and 2Q is the diagonal of the rectagle of length l:

$\displaystyle |d_3|=\sqrt{l^2+l^2}=\sqrt{2}\ l$

The force F3 is

$\displaystyle |F_3|=\frac{K(3Q)(2Q)}{(\sqrt{2l)}^2}=\frac{3KQ^2}{l^2}$

Each force must be expressed as vectors. F1 is pointed to the right direction, thus its vertical components is zero

$\displaystyle \vec{F_1}=\left \langle |F_1|,0 \right \rangle=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle$

F2 is pointed upwards and its horizontal component is zero

$\displaystyle \vec{F_2}=\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle$

F3 has two components because it forms an angle of 45° respect to the horizontal, thus

$\displaystyle \vec{F_3}=\left \langle \frac{3KQ^2}{l^2}\ cos45^o,\frac{3KQ2}{l^2} sin45^o\right \rangle$

$\displaystyle \vec{F_3}=\left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

Now we compute the total force

$\displaystyle \vec{F_t}=\vec{F_1}+\vec{F_2}+\vec{F_3}$

$\displaystyle \vec{F_t}=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle +\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle + \left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

$\displaystyle \vec{F_t}=\left \langle \left(2+\frac{3\sqrt{2}}{2}\right)\frac{KQ^2}{l^2},\left(8+\frac{3\sqrt{2}}{2}\right) \frac{KQ^2}{l^2}\right \rangle$

$\displaystyle F_t=\left \langle 4.121,10.121 \right \rangle \frac{KQ^2}{l^2}$

Now we compute the magnitude

$\boxed{\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}}$

The direction of the total force is given by

$\displaystyle tan\theta =\frac{10.121}{4.121}=2.4558$

$\boxed{\displaystyle \theta =68^o}$

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3 years ago