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2 years ago

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You send a traveling wave along a particular string by oscillating one end. If you increase the frequency of oscillations, does

the speed of the wave increase, decrease, or remain the same?

1 answer:

eimsori [14]2 years ago

4 0

Answer:

The speed of the wave remains the same

Explanation:

Since the speed of the wave v = √(T/μ) where T is the tension in the string and μ is the linear density of the string.

We observed that the speed, v is independent of the frequency of the wave in the string. So, increasing the frequency of the wave has no effect on the speed of the wave in the string, since the speed of the wave in the string is only dependent on the properties of the string.

<u>So, If you increase the frequency of oscillations, the speed of the wave remains the same.</u>

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An empty parallel plate capacitor is connected between the terminals of a 18.8-V battery and charges up. The capacitor is then d

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Answer:

$p.d' = 37.6 V$

Explanation:

From the question we are told that:

Potential difference $p.d=18.8V$

New Capacitor $C_1=C_2/2$

Generally the equation for Capacitor capacitance is mathematically given by

$C=\frac{eA}{d}$

Generally the equation for New p.d' is mathematically given by

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$p.d' = 2V$

$p.d'= 2 * 18.8$

$p.d' = 37.6 V$

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The answer is c. interest.

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Gamma rays have a higher frequency than visible light waves. What can you conclude from this?

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3 years ago

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Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into th

satela [25.4K]

The tank has a volume of $\dfrac\pi3R^2H$ , where $H=6\,\rm m$ is its height and $R=\dfrac d2=2\,\rm m$ is its radius.

At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:

$\dfrac26=\dfrac rh\implies r=\dfrac h3$

The volume of water in the tank at any given time is

$V=\dfrac\pi3r^2h$

and can be expressed as a function of the water level alone:

$V=\dfrac\pi3\left(\frac h3\right)^2h=\dfrac\pi{27}h^3$

Implicity differentiating both sides with respect to time $t$ gives

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9h^2\,\dfrac{\mathrm dh}{\mathrm dt}$

We're told the water level rises at a rate of $\dfrac{\mathrm dh}{\mathrm dt}=20\,\frac{\rm cm}{\rm min}$ at the time when the water level is $h=2\,\mathrm m=200\,\mathrm{cm}$ , so the net change in the volume of water $\dfrac{\mathrm dV}{\mathrm dt}$ can be computed:

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9(200\,\mathrm{cm})^2\left(20\,\dfrac{\rm cm}{\rm min}\right)=\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}$

The net rate of change in volume is the difference between the rate at which water is pumped into the tank and the rate at which it is leaking out:

$\dfrac{\mathrm dV}{\mathrm dt}=(\text{rate in})-(\text{rate out})$

We're told the water is leaking out at a rate of $10,500\,\frac{\mathrm{cm}^3}{\rm min}$ , so we find the rate at which it's being pumped in to be

$\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}=(\text{rate in})-10,500\,\dfrac{\mathrm{cm}^3}{\rm min}$

$\implies\text{rate in}\approx289,753\,\dfrac{\mathrm{cm}^3}{\rm min}$

4 0

3 years ago

A 4.00-kg block hangs by a light string that passes over a massless, frictionless pulley and is connected to a 6.00-kg block tha

Anna007 [38]

Answer:

v = 2.82 m/s

Explanation:

For this exercise we can use the conservation of energy relations.

We place our reference system at the point where block 1 of m₁ = 4 kg

starting point. With the spring compressed

Em₀ = K_e + U₂ = ½ k x² + m₂ g y₂

final point. When block 1 has descended y = - 0.400 m

Em_f = K₂ + U₂ + U₁ = ½ m₂ v² + m₂ g y₂ + m₁ g y

as there is no friction, the energy is conserved

Em₀ = Em_f

½ k x² + m₂ g y₂ = ½ m₂ v² + m₂ g y₂ + m₁ g y

½ k x² - m₁ g y = ½ m₂ v²

v² = $\frac{k}{m_2} x^2 - 2 \frac{m_1}{m_2} \ g y$

let's calculate

v² = $\frac{180}{6.00} \ 0.300^2 - 2 \ \frac{4.00}{6.00} \ 9.8 \ (- 0.400)$

v² = 2.7 + 5.23

v = √7.927

v = 2,815 m / s

using of significant figures

v = 2.82 m/s

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3 years ago