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3 years ago

Complete the passage to differentiate between longitudinal and transverse waves.

Physics

2 answers:

koban [17]3 years ago

8 0

Answer:

parallel, and perpendicular

Explanation:

astra-53 [7]3 years ago

3 0

Complete sentence:

A longitudinal wave is a type of wave that transfers energy parallel to the direction of wave motion. A transverse wave, on the other hand, is a type of wave that transfers energy perpendicular to the direction of wave motion.

Explanation

the main difference between longitudinal and transverse wave is the direction of their oscillation/vibration. In a longitudinal wave, the oscillation of the wave occurs in a direction parallel to the direction of propagation of the wave, while in a transverse wave, the oscillation occurs in a plane perpendicular to the direction of propagation. An example of transverse waves are the electromagnetic waves, which consists of an electric field and a magnetic field that oscillates perpendicular to the direction of the wave, while an example of longitudinal waves are the sound waves, whose particles vibrate in a direction parallel to the direction of motion of the wave, producing regions of more density of particles (called compressions) and regions of less density of particles (called rarefactions).

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An output gear has 10 teeth and an input gear has 40 teeth. What is the mechanical advantage of gear combination? Is force or sp

yKpoI14uk [10]

Answer:

Mechanical advantage: 4. Speed is multiplied

Explanation:

The mechanical advantage of gear combination is

$MA = \frac{N_i}{N_o}$

where

$N_i$ is the number of teeth in the input gear

$N_o$ is the number of teeth in the output gear

Here we have

$N_i = 40\\N_o = 10$

So we find

$MA = \frac{40}{10}=4$

In this example, it is the speed to be multiplied: the speed of the output gear is 4 times larger than the speed of the input gear. In fact, the mechanical advantage can be rewritten as

$MA= \frac{\omega_o}{\omega_i}$

where

$\omega_i$ is the angular speed of the input gear

$\omega_o$ is angular speed of the output gear

So re-arranging the equation, we find

$\omega_o = MA \cdot \omega_i = 4 \omega_i$

so, the speed of the output gear is 4 times larger than the speed of the input gear.

7 0

3 years ago

Sound travels at a rate of 340 m/s in all directions through air. Matt rings a very loud bell at one location, and Steve hears i

Mice21 [21]

It took 450/340 = 1.32 seconds

5 0

4 years ago

What is the focal length of concave mirror that magnifies , by a factor of +3.2 , an object that is placed 30cm from the mirror?

guapka [62]

Answer:

23 cm

Explanation:

The formula for magnification is;

Magnification = image distance / object distance

use values as;

3.2 = v / 30 where v is image distance

v =30*3.2

v=96 cm

The relationship of the focal length with image distance and object distance is expressed as;

$\frac{1}{u} +\frac{1}{v} =\frac{1}{f}$

where f is the focal length and u is object distance

use values in the equation as;

$\frac{1}{30} +\frac{1}{96 } =\frac{1}{f}$

$\frac{1}{f} =\frac{7}{160}$

f=160/7

f=22.86

f= 23 cm ----------nearest a cm

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3 years ago

A spiral spring has a length of 14 cm when a force of 4 N is hung on it. A force of 6 N extends

stira [4]

Answer:

bussy

Explanation:

shart

5 0

2 years ago

A thin, uniformly charged insulating rod has a linear charge density λ = 3 nC/m and lies along the x axis from x = 1m to x = 3m.

Contact [7]

Answer:

A) $V_A = 11.93~V$

B) The vector definition of E-field is

$\vec{E} = -1.13\^x + 2.41\^y$

where magnitude is E = 2.66 N/m.

Explanation:

The potential of a uniformly charged rod can be found by the method of integration. We will first choose an infinitesimal part on the rod. We will compute the potential of this part at point A. Then we will integrate this potential over the entire rod.

We will use the following formula for electric potential:

$V = \frac{1}{4\pi \epsilon_0}\frac{Q}{r}$

Let us choose the infinitesimal part a distance 'x' from the origin. Then the distance between this point and point A is

$r = \sqrt{x^2+4^2}$

The infinitesimal length is 'dx', and the potential of this length is dV. Let's apply the formula:

$dV = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{\sqrt{x^2 + 4^2}}$

Here, the charge Q is equal to the charge density multiplied by the length. Q = λdx

Now we have to integrate this infinitesimal potential over the rod:

$V = \int\limits^3_1 {dV} \, dx = \frac{1}{4\pi \epsilon_0}\int\limits^3_1 {\frac{\lambda}{\sqrt{x^2 + 16}} \, dx$

By using an integral table, this can be calculated:

$V = \frac{3\times 10^{-9}}{4\pi\epsilon_0}\ln(|\sqrt{x^2+16}+x|)\left \{ {{x=3} \atop {x=1}} \right. \\V = 11.93~V$

B) The electric field can be found by a similar approach, but a different formula:

$\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\^r$

Let's apply this formula to the infinitesimal part we have chosen.

$dE_x = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\cos(\theta)\\dE_y = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\sin(\theta)$

By the geometry sine and cosine terms can be found:

$\sin(\theta) = \frac{4}{\sqrt{x^2+16}}\\\cos(\theta) = \frac{x}{\sqrt{x^2 + 16}}$

The x- and y-components of the E-field can be found separately by integrating the infinitesimal parts over the entire rod.

$E_x = \int\limits^3_1 {dE_x} \, dx = \frac{\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{x}{(x^2+16)^{3/2}}} \, dx = 1.13(-\^x)\\E_y = \int\limits^3_1 {dE_y} \, dx = \frac{4\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{1}{(x^2+16)^{3/2}}} \, dx = 2.41(\^y)$

So, the final E-field is

$\vec{E} = -1.13\^x + 2.41\^y$

The magnitude of the E-field is

E = 2.66 N/m

6 0

3 years ago