All categories

3 years ago

The major difference between transverse and longitudinal waves is the:

1 answer:

3 0

In transversal wave particles do not oscillate along the line of the wave propagation but oscillate up and down about their mean position as the wave travels.thus transverse wave is the one in which individual particles of the medium move about their mean position in a direction perpendicular to the direction of wave propprgation.
In longitutional waves the individual particles of the medium move in the direction parallel to the direction of the propprgation of the disturbance. The particles do not move from one place to another but the simply oscillate back and forth about their position of the rest.
HOPE IT HELPS.

You might be interested in

If a cylindrical space station 275 m in diameter is to spin about its central axis, at how many revolutions per minute (rpm) mus

sergij07 [2.7K]

The centripetal acceleration is responsible for the artificial gravity because the acceleration of an object moving in constant circular motion causing from net external force is called centripetal acceleration. It defines to the center or seeking the center.

Given the following:

Cylindrical space station diameter = 275 meters; 137.5 meters for the radius

Standard gravity = 9.80665 m/s²

Using the formula:

w² x r =g

w² = g / r

w² = 9.80665 m/s² / 137.5 m

w² = 0.0713 s²

Then take the roots

w = 0.267 this is radians per second / 2 x (3.1416 which is the pi)

w = 0.0424 rps convert to rpm

w = 0.0424 r/s (1minute / 60 seconds)

w = 7.08 x 10⁻⁴ revolutions per minute

4 0

3 years ago

A 1000kg car is rolling slowly across a level surface at 1 m/s heading twoards a group o fsmall innocent children. The doors are

Degger [83]

Answer:

The force required to push to stop the car is 288.67 N

Explanation:

Given that

Mass of the car, m = 1000 kg

Initial speed of the car, u = 1 m/s

The car and push on the hood at an angle of 30° below horizontal, $\theta=30^{\circ}$

Distance, d = 2 m

Let F is the force must you push to stop the car.

According work energy theorem theorem, the work done is equal to the change in kinetic energy as :

$W=\dfrac{1}{2}m(v^2-u^2)F\times d=\dfrac{1}{2}m(v^2-u^2)$

$v = 0$

$Fd\ cos\theta=\dfrac{1}{2}m(u^2) F=\dfrac{\dfrac{1}{2}m(u^2)}{d\ cos\theta}F=\dfrac{\dfrac{1}{2}\times 1000\times (1)^2}{2\ cos(30)}F = -288.67 N$

The force required to push to stop the car is 288.67 N

3 0

2 years ago

A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos

lukranit [14]

Answer:

The inductor contains $N = 523962.32$ loops

Explanation:

From the question we are told that

The capacitance of the capacitor is $C = 286nF = 286 * 10^{-9} \ F$

The resonance frequency is $f = 18.0 kHz = 18*10^{3} Hz$

The diameter is $d = 1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m$

The of the air-core inductor is $l = 12 \ m$

The permeability of free space is $\mu_o = 4 \pi *10^{-7} \ T \cdot m/A$

Generally the inductance of this air-core inductor is mathematically represented as

$L = \frac{\mu_o * N^2 \pi d^2}{4 l}$

This inductance can also be mathematically represented as

$L = \frac{1}{w^2}$

Where $w$ is the angular speed mathematically given as

$w = 2 \pi f$

$L = \frac{1}{4 \pi ^2 f^2}$

Now equating the both formulas for inductance

$\frac{\mu_o * N^2 \pi d^2}{4 l} = \frac{1}{4 \pi ^2 f^2}$

making N the subject of the formula

$N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C} }$

$N = \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }$

Substituting value

$N = \frac{1}{ 3.142 * 18*10^{3} * 0.00011 } \sqrt{\frac{12}{ 3.142 * 4 \pi *10^{-7}* 286 *10^{-9}} }$

$N = 523962.32$ loops

4 0

3 years ago

A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut

blsea [12.9K]

Answer:

1. $E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. $E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

$\rho (r) = \alpha (1-\frac{r}{R})$

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

$\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

$\int\, da = 2\pi r h$

where ‘h’ is the length of the imaginary Gaussian surface.

$Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

$Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3. At the boundary where r = R:

$E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}$

As can be seen from above, two E-field values are equal as predicted.

4 0

3 years ago

Suppose we repeat the experiment from the video, but this time we use a rocket three times as massive as the one in the video, a

shusha [124]

Answer:

2/3

Explanation:

In the case shown above, the result 2/3 is directly related to the fact that the speed of the rocket is proportional to the ratio between the mass of the fluid and the mass of the rocket.

In the case shown in the question above, the momentum will happen due to the influence of the fluid that is in the rocket, which is proportional to the mass and speed of the same rocket. If we consider the constant speed, this will result in an increase in the momentum of the fluid. Based on this and considering that rocket and fluid has momentum in opposite directions we can make the following calculation:

Rocket speed = rocket momentum / rocket mass.

As we saw in the question above, the mass of the rocket is three times greater than that of the rocket in the video. For this reason, we can conclude that the calculation should be done with the rocket in its initial state and another calculation with its final state:

Initial state: Speed = rocket momentum / rocket mass.

Final state: Speed = 2 rocket momentum / 3 rocket mass. -------------> 2/3

8 0

3 years ago