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3 years ago

If a 4.5-meter wave traveled at a speed of 15 m/s, what would its frequency be?

Physics

2 answers:

kolezko [41]3 years ago

5 0

Answer:

The frequency of the wave = 10 Hz.

Explanation:

Wave: A wave is a disturbance that travels through a medium a transfer energy from one point to another in the medium without causing and permanent displacement of the medium itself.

V = λf .................. Equation 1

making f the subject of the equation,

f = V/λ.................... Equation 2

Where V = velocity of the wave, λ = wavelength of the wave, f = frequency of the wave.

Given: V = 15 m/s², λ = 4.5 m.

Substituting these values into equation 2,

f = 15/1.5

f = 10 Hz.

Therefore the frequency of the wave = 10 Hz.

Setler [38]3 years ago

3 0

Answer:

The frequency of the wave = 10 Hz.

Explanation:

the guy below gave a good explanation

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Explanation:

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3 years ago

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Rashid [163]

Answer:

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Explanation:

Solution:-

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4 years ago

3% of earth's water is?

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3 years ago

Two parallel-plate capacitors have the same dimensions, but the space between the plates is filled with air in capacitor 1 and w

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Explanation:

The capacitance in two parallel-plate capacitors is:

$C=\frac{K\epsilon_oA}{d}$

For air, we have $K_1=1$

For plastic, we have $K_2=2.25$

Hence:

$C_1=\frac{K_1\epsilon_oA}{d_1}=\frac{\epsilon_oA}{d_1}\\C_2=\frac{K_2\epsilon_oA}{d_2}=2.25(\frac{\epsilon_oA}{d_2})$

a) Recall that the potential difference between the plates is the same ( $V_1=V_2=V$ ). The electric field is given by:

$E_1=\frac{V_1}{d_1}=\frac{V}{d_1}\\E_2=\frac{V_2}{d_2}=\frac{V}{d_1}$

The potential difference is defined as:

$V_1=\frac{Q_1}{C_1}=V\\V_2=\frac{Q_2}{C_2}=V$

Replacing:

$E_1=\frac{Q_1}{C_1d_1}\\E_2=\frac{Q_1}{C_1d_2}\\\\E_1d_1=\frac{Q_1}{C_1}\\E_2d_2=\frac{Q_1}{C_1}\\E_1d_1=E_2d_2\\\frac{E_1}{E_2}=\frac{d_2}{d_1}$

b) The energy is defined as:

$U=\frac{1}{2}CV^2$

So:

$U_1=\frac{1}{2}C_1V_1^2=\frac{1}{2}C_1V^2\\U_2=\frac{1}{2}C_2V_2^2=\frac{1}{2}C_2V^2\\U_1=\frac{1}{2}\frac{\epsilon_oA}{d_1}V^2\\U_2=\frac{1}{2}\frac{2.25\epsilon_oA}{d_2}V^2\\\frac{0.88U_2d_2}{\epsilon_oA}=V^2\\\frac{2U_1d_1}{\epsilon_oA}=V^2\\\frac{0.88U_2d_2}{\epsilon_oA}=\frac{2U_1d_1}{\epsilon_oA}\\\frac{U_1}{U_2}=0.44\frac{d_2}{d_1}$

5 0

3 years ago

A portable basketball set has a base and a post arrangement. The post arrangement consists of a post, backboard, hoop and net. T

Ray Of Light [21]

The rotational equilibrium condition allows finding the response to the minimum force of the wind and what happens when changing the water for sand, in the system

a) The minimum force of the wind that turns the system is Fw = 17.64 N

b) The system resists much greater forces because the base has more mass

Newton's Second Law can be applied to rotational motion in this case when the angular acceleration is zero we have the special case of rotational equilibrium

Σ τ = 0

Where τ is the torque

The reference system is a coordinate system with respect to which the torques are measured, in this case we will fix the system at the turning point, the junction of the base and the pole, we will assume that the counterclockwise rotations are positive.

For the torque the distance used is the perpendicular distance from the direction of the force to the axis of rotation, let's find this distance for each force

Wind force

cos 15 = $\frac{y_w}{2.35}$

$y_w$ = 2.35 cos 15

Post Weight

sin 15 = $\frac{x_p}{2.00}$

xp = 2.0 sin 15

Base weight

cos (90-15) = $\frac{x_b}{0.25}$

xB = 0.25 cos 75

Let's substitute in the rotational equilibrium equation

$F_w \ y_w + W_p \ x_p - W_b \ x_b = 0$

a) To calculate the minimum wind force we substitute the given values

They indicate the weight of the post is $W_p$ = 26.0 N and the weight of the base with water is $W_b$ = 810 N

$F_w = \frac{W_b \ x_b - W_p \ x_p }{y_w}$

$F_w = \frac{W_b \ 0.25 cos75 \ - W_p \ 2 sin 15}{2.35 cos 15}$

Let's calculate

$F_w = \frac{810 \ 0.25 \ cos75 \ - 26.0 \ 2 \ sin 15}{2.35 cos15}\\F_w = \frac{52.41 - 10.30}{2.3699}$

$F_w$ = 17.64 N

b) The water is exchanged for sand.

In this case, as the density of the sand is greater than that of the water, the base will have more weight, so it will resist stronger winds before turning over.

Using the rotational equilibrium condition we can find the response to the minimum force of the wind and what happens when changing the water for sand,

a) the minimum force of the wind that turns the system is Fw = 17.64 N

b) the system resists much greater forces because the base has more mass

Learn more here: brainly.com/question/7031958

3 0

3 years ago