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4 years ago

6

When a tuning fork vibrates over an open pipe and the air in the pipe starts to vibrate, the vibrations in the tube are caused b

y A. resonance. B. beats. C. harmonics. D. reinforcement.

2 answers:

Fed [463]4 years ago

6 0

Letter A would be your best awnser for this question

Likurg_2 [28]4 years ago

5 0

This is caused by A. resonance.
hope this helps!

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Which is larger, the Sun's pull on Earth or Earth's pull on the Sun?

Juliette [100K]

B it’s the most logical answer

3 0

4 years ago

At its closest approach, a moon comes within 200,000 km of the planet it orbits. At that point, the moon is 300,000 km from the

krok68 [10]

Answer:

s₁ = 240,000 km

Explanation:

The distance between both the focuses f₁ and f₂ will be the sum of distances of the moon from each focus at a given point. Therefore,

s = s₁ + s₂

where,

s = total distance between the focuses = ?

s₁ = distance between f1 and moon = 200,000 km

s₂ = distance between f₂ and moon = 300,000 km

Therefore,

s = 200,000 km + 300,000 km

s = 500,000 km

Now, when the distance from f₂ becomes 260,000 km, then the distance from f₁(planet) will become:

s = s₁ + s₂

500,000 km = s₁ + 260,000 km

s₁ = 500,000 km - 260,000 km

<u>s₁ = 240,000 km</u>

5 0

3 years ago

A long, thin solenoid has 450 turns per meter and a radius of 1.17 cm. The current in the solenoid is increasing at a uniform ra

sergey [27]

Answer:

$\frac{di}{dt} = 7.31 \ A/s$

Explanation:

From the question we are told that

The number of turns is $N = 450 \ turns$

The radius is $r = 1.17 \ cm = 0.0117 \ m$

The position from the center consider is x = 3.45 cm = 0.0345 m

The induced emf is $e = 8.20 *10^{-6} \ V/m$

Generally according to Gauss law

$\int\limits { e } \, dl = \mu_o * N * \frac{di}{dt } * A$

=> $e * 2\pi x = \mu_o * N * \frac{d i }{dt } * A$

Where A is the cross-sectional area of the solenoid which is mathematically represented as

$A = \pi r ^2$

=> $e * 2\pi x = \mu_o * N * \frac{d i }{dt } * \pi r^2$

=> $\frac{di}{dt} = \frac{2e * x }{\mu_o * N * r^2}$ ggl;

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} \ N/A^2$

=> $\frac{di}{dt} = \frac{2 * 8.20*10^{-6} * 0.0345 }{ 4\pi * 10^{-7} * 450 * (0.0117)^2}$

=> $\frac{di}{dt} = 7.31 \ A/s$

6 0

4 years ago

A student measures the diameter of a small cylindrical object and gets the following readings: 4.32, 4.35, 4.31, 4.36, 4.37, 4.3

Zinaida [17]

Answer:

a. $\bar{d}=4.34 cm$

b. $\sigma=0.023 cm$

c. $\rho=(0.0089\pm 0.00058) kg/cm^{3}$

Explanation:

a) The average of this values is the sum each number divided by the total number of values.

$\bar{d}=\frac{\Sigma_{i=1}^(N)x_{i}}{N}$

$x_{i}$ is values of each diameter
N is the total number of values. N=6

$\bar{d}=\frac{4.32+4.35+4.31+4.36+4.37+4.34}{6}$

$\bar{d}=4.34 cm$

b) The standard deviation equations is:

$\sigma=\sqrt{\frac{1}{N}\Sigma^{N}_{i=1}(x_{i}-\bar{d})^{2}}$

If we put all this values in that equation we will get:

$\sigma=0.023 cm$

Then the mean diameter will be:

$\bar{d}=(4.34\pm 0.023)cm$

c) We know that the density is the mass divided by the volume (ρ = m/V)

and we know that the volume of a cylinder is: $V=\pi R^{2}h$

Then:

$\rho=\frac{m}{\pi R^{2}h}$

Using the values that we have, we can calculate the value of density:

$\rho=\frac{1.66}{3.14*(4.34/2)^{2}*12.6}=0.0089 kg/cm^{3}$

We need to use propagation of error to find the error of the density.

$\delta\rho=\sqrt{\left(\frac{\partial\rho}{\partial m}\right)^{2}\delta m^{2}+\left(\frac{\partial\rho}{\partial d}\right)^{2}\delta d^{2}+\left(\frac{\partial\rho}{\partial h}\right)^{2}\delta h^{2}}$

δm is the error of the mass value.
δd is the error of the diameter value.
δh is the error of the length value.

Let's find each partial derivative:

1. $\frac{\partial\rho}{\partial m}=\frac{4m}{\pi d^{2}h}=\frac{4*1.66}{\pi 4.34^{2}*12.6}=0.0089$

2. $\frac{\partial\rho}{\partial d}=-\frac{8m}{\pi d^{3}h}=-\frac{8*1.66}{\pi 4.34^{3}*12.6}=-0.004$

3. $\frac{\partial\rho}{\partial h}=-\frac{4m}{\pi d^{2}h^{2}}=-\frac{4*1.66}{\pi 4.34^{2}*12.6^{2}}=-0.00071$

Therefore:

$\delta\rho=\sqrt{\left(0.0089)^{2}*0.05^{2}+\left(-0.004)^{2}*0.023^{2}+\left(-0.00071)^{2}*0.5^{2}}$

$\delta\rho=0.00058$

So the density is:

$\rho=(0.0089\pm 0.00058) kg/cm^{3}$

I hope it helps you!

3 0

3 years ago

Which one of the following types of electromagnetic radiation causes certain substances to fluoresce?

ioda

From the choices given, ultraviolet rays is a type electromagnetic radiation that causes certain substances to fluoresce. It cannot be detected by the naked eye but some insects are able to see them.

Like in a fluorescent light bulb, ultraviolet or UV lights stimulates the coating of the tube to emit light.

5 0

3 years ago