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3 years ago

12

This chart shows characteristics of three different waves, all with the same wavelength of 10 m but moving at different Which st

atement is best supported by the information in the chart?
A 3-column table with 2 rows titled Waves at Different Frequencies. The first column has entries Wave X and 6 hertz. The second column has entries Wave Y and 0.5 hertz. The third column has entries Wave Z and 2 hertz.
Wave Y is moving the fastest.

Wave X is moving the fastest.

All of the waves are moving at the same speed.

All of the waves have the same value of wavelengths per second.

2 answers:

Ainat [17]3 years ago

7 0

Answer:

Wave X is moving the fastest.

JulijaS [17]3 years ago

3 0

Answer:

Wave X is moving the fastest.

Explanation:

As we know that the speed of the wave is given as the product of frequency and wavelength

so here we will have

$\lambda = 10 m$

now for wave X

$f = 6 Hz$

now the speed of this wave is

$v_x = 10 \times 6 = 60 m/s$

now for wave Y

$f = 0.5 Hz$

now the speed of this wave is

$v_y = 10 \times 0.5 = 5 m/s$

now for wave Z

$f = 2 Hz$

now the speed of this wave is

$v_z = 10 \times 2 = 20 m/s$

So correct answer will be

Wave X is moving the fastest.

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A dog is walking back home and its motion is modeled by the graph above. What position was the dog at 3 seconds after it started

hammer [34]

6 meters because 6/3 = 10/5

4 0

3 years ago

Three charges are located at a different position in a plane: q1= 10μC at →r1=(5,6)cm q2=−27μC at →r2=(−6,10)cm and q3=−12μC at

sasho [114]

Answer:

E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

E_{total} = 2,467 10⁶ N / A θ = -21.8

Explanation:

For this exercise we will use that the electric field is a vector quantity, so the total field is

E_total = E₁₃ + E₂₃

bold font vectors . We can work with the components of the electric field in each axis

X- axis

E_ total x = E₁₃ₓ + E_{23x}

y-axis

E_{total y} = E_{13y} + E_{23y}

the expression for the electric field is

E = k q / r²

where r is the distance between the charge and the positive test charge

in this exercise

Let's find the field created by charge 1

q₁ = 10 μC = 10 10⁻⁶ C

x₁ = 5 cm = 0.05 m

x₃ = 21 cm = 0.21 m

E_{13x} = 9 10⁹ 10 10⁻⁶ / (0.21 -0.05)²

E_{13x} = 3.516 10⁶ N / C

y₁ = 6 cm = 0.06 cm

y₃ = -12 cm = -0.12 m

E_{13y} = 9 10⁹ 10 10⁻⁶ / (-0.12 - 0.06)²

E_{13y} = 2,777 10⁶ N / C

let's find the field produced by charge 2

q₂ = -27 μC = - 27 10⁻⁶ C

x₂ = -6 cm = -0.06 m

x₃ = 0.21 m

E_{23x} = 9 10⁹ 27 10⁻⁶ / (0.21 + 0.06)²

E_{23x} = 1.23 10⁶ N / A

y₂ = 10 cm = 0.10 m

y₃ = -0.12 m

E_{23y} = 9 10⁹ 27 10⁻⁶ / (-0.12 - 0.10)²

E_{23y} = 1.86 10⁶ N / C

Taking the components we can calculate the total electric field, we must use that charge of the same sign repel and attract the opposite sign, remember that the test charge is always considered positive.

E_{total x} = E_{13x} - E_{23x}

E_{total x} = (3.516 - 1.23) 10⁶

E_{total x} = 2.29 10⁶ N / A

E_{total y} = -E_{13y} + E_{23y}

E_{total y} = (-2.777 +1.86) 10⁶ N / A

E_{total y} = -0.917 10⁶ N / A

we can give the result in two ways

E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

or in the form of modulus and angle, let's use the Pythagorean theorem to find the modulus

E_{total} = √ (E_{total x}^2 + E_{total y}^ 2)

E_{total} = √ (2.29² + 0.917²) 10⁶

E_{total} = 2,467 10⁶ N / A

let's use trigonometry for the angle

tan θ = E_total and / E_totalx

θ = tan⁻¹ E_{total y} / E_{total x}

θ = tan⁻¹ (-0.917 / 2.29)

θ = -21.8

The negative sign indicates that the angle is measured with respect to the x-axis in a clockwise direction.

7 0

3 years ago

In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir

Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be $S_{1}$

and the distance covered by Sir Alfred be $S_{2}$

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

$S = ut + \frac{1}{2}at^{2}$

Where $S$ is the distance traveled

$u$ is the initial velocity

$a$ is the acceleration

and $t$ is the time

For Sir George,

$S = S_{1}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.21 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{1} = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1} = 0.105 t^{2}\\$

$t^{2} = \frac{S_{1}}{0.105}$

Now, for Sir Alfred

$S = S_{2}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.26 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{2} = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2} = 0.13 t^{2}\\$

$t^{2} = \frac{S_{2}}{0.13}$

Since, they traveled for the same time, $t$ just before collision, we can write

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

$S_{1} + S_{2} = 86$ m

Then, $S_{2} = 86 - S_{1}$

Then,

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$ becomes

$\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}$

$0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}$

$S_{1} = 38.43$ m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

6 0

3 years ago

Why does the picture on a tv screen become distorted when a magnet is brought near the screen<br>

NemiM [27]

When a magnet is brought close to the picture tube, the interaction between the flying electrons and the magnetic field creates a force that throws the electrons off course. Now the electrons are hitting the screen in places they were not intended to strike and the picture becomes distorted.

When a television receives a signal, it first splits off the audio (sound) signal and the picture signal from a carrier wave (which is used to allow the signal to be transmitted over long distances). The audio is sent straight to the speakers to produce sound. The picture signal consists of three elements, red, green and blue. A standard television has three 'electron guns' at the back of the set, one for each colour. Let's start by looking at the red signal. The red signal is fed into one of these 'guns'. The gun produces a beam of electrons that varies in intensity with the strength of the red signal. This beam is fired towards the tv screen. The electron beam starts at the top-left of the screen and magnetic fields are used to 'sweep' this beam across the screen in parallel horizontal lines (if you look closely at a tv screen you can see these lines). UK televisions (PAL) have 625 lines and update the picture 25 times per second, US televisions (NTSC) have 525 lines but update 30 times per second. The back of the tv screen is covered in phosphor 'dots' (pixels) which glow when they are struck by these electrons. The red-signal electron beam is aimed so that it strikes phosphor dots that glow red, emitting photons which the eye can detect. The same process occurs for green and blue; each colour signal goes to one particular electron gun which excites just the dots of that colour, the signal tells the gun how strong it should be which in turn means some dots glow brighter than others. When you sit back from the tv screen, you don't notice the dots nor the flicker, your eye blends the image together to give a clear picture which appears to move. Now to answer the question! A magnet distorts the picture as it distorts the path of electrons flowing from the electron gun towards the screen inside the tv. As electrons are negatively charged particles, their motion is distorted by a magnet. So it is these electrons, not photons, which are distorted by the magnet. On older tvs, damage caused by holding a magnet too close to a tv could be permanent; newer tvs tend to have a demagnetisation process when you switch them on, to ensure that the picture is not permanently distorted. ehehe..

3 0

3 years ago

A rectangular coil of 65 turns, dimensions 0.100 m by 0.200 m, and total resistance 10.0 ? rotates with angular speed 29.5 rad/s

vovikov84 [41]

Answer:

Explanation:

N = 65

Area, A = 0.1 x 0.2 = 0.02 m^2

R = 10 ohm

ω = 29.5 rad/s

B = 1 T

(a) at t = 0

e = N x B x A x ω

e = 65 x 1 x 0.02 x 29.5

e = 38.35 V

(b) The maximum rate of change of magnetic flux is equal to the maximum value of induced emf.

Ф = 38.35 Wb/s

(c) e = NBAω Sinωt

e = 65 x 1 x 0.02 x 29.5 x Sin (29.5 x 0.05)

e = 38.174 V

(d) Maximum torque

τ = M B Sin 90

τ = N i A B

τ = N e A B / R

τ = 65 x 38.35 x 0.02 x 1 / 10

τ = 5 Nm

8 0

3 years ago