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3 years ago

15

As a wave travels through a medium, it displaces particles in a direction parallel to the motion of the wave. We can conclude th

is is a

2 answers:

SIZIF [17.4K]3 years ago

8 0

This is a longitudinal wave.
Any wave traveling through liquid or gas does it this way.
An example is: Sound waves.

olga nikolaevna [1]3 years ago

7 0

longitudinal wave because particle displacement is the same direction as the wave

hope this helps :/

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State keplers law........

S_A_V [24]

Answer:

There are actually three, Kepler's laws that is, of planetary motion: 1) every planet's orbit is an ellipse with the Sun at a focus; 2) a line joining the Sun and a planet sweeps out equal areas in equal times; and 3) the square of a planet's orbital period is proportional to the cube of the semi-major axis of its

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3 years ago

Which of the following scenarios would be optimal for obtaining a date from radioactive decay using these isotopes: 87Rb, 147Sm,

REY [17]

Answer:

a) 238U, 40K and 87Rb, b) 235U and to a lesser extent 40K , c) he 235U,

d) possibility is 14C , e)this period would be ideal for 14C , f) 14C should be used since it is the one with the least average life time, even though the measurements must be very careful

Explanation:

One of the applications of radioactive decay is the dating of different systems.

To do this, the quantity of radioactive material in a meter is determined and with the average life time, the time of the sample is found.

Let's write the half-life times of the given materials

87Rb T ½ = 4.75 1010 years

147Sm T ½ = 1.06 1011 years

235U = 7,038 108 years

238U = 4.47 109 years

40K = 1,248 109 years

14C = 5,568 103 years

we already have the half-life of the different elements given

a) meteors. As these decomposed in the formation of the solar system, their life time is around 3 109 to 5 109 years, so it is necessary to look for elements that have a life time of this order, among the candidates we have 238U, 40K and 87Rb if these elements were at the moment of the formation of these meteors, there must still be rations in them, instead elements 14C already completely adequate

b) rock. The formation period is 4.20-108 years, therefore one of the most promising elements is 235U and to a lesser extent 40K since it is more abundant in rocks. The other elements with higher life times have not decayed and therefore will not give a true value and the 14C is completely decayed

c) volcanic ash. Formation time 6107 years, the only element that has the possibility of having a count is the 235U, the others have a life time so long that they have not decayed and the 14C is complete, unbent

d) scarp of an earthquake formation time 5 101 years, The only one that has any possibility is 14C even when it has declined very little, all the others, you have time to long that has not decayed

e) INCA excavation. The time of this civilization is about 10000 to 500 years (104 to 5 102 years), we see that this period would be ideal for 14C since it has some period of cementation, the others have not decayed

f) Tree in Blepharitis. 14C should be used since it is the one with the least average life time, even though the measurements must be very careful because of a period of disintegration. We have such a long time that they have not decayed

8 0

4 years ago

A listener is sitting somewhere on the line between two loudspeakers that are 10 m apart. The speakers are each emitting a sine

prohojiy [21]

Answer:

57.17 Hz 114.34 Hz 285 Hz

Explanation:

The distance between the men and 1 speaker = 3.5 m

Distance between the men and second speaker = 10-3.5= 6.5 m

Here at this point there will be no sound so there will be destructive interference

Path difference $\Delta x=6.5-3.5=3$

We know that for destructive interference $\Delta x=(2m+1)\frac{\lambda }{2}=(2m+1)\frac{v}{2f}$

$3=(2m+1)\frac{v}{2f}$

$f=(2m+1)\frac{v}{6}$ here v is the speed of sound in air

So for m =0

$f=(2\times 0+1)\frac{343}{6}=57.17H$

for m =1

$f=(2\times 1+1)\frac{343}{6}=114.34Hz$

for m=2

$f=(2\times 2+1)\frac{343}{6}=285Hz$

6 0

3 years ago

Two charged concentric spheres have radii of 0.008 m and 0.018 m. The charge on the inner sphere is 3.62 10-8 C and that on the

Alekssandra [29.7K]

Answer:

The electric field is $E = 2.2625 *10^{6} \ N/C$

Explanation:

From the question we are told that

The radius of the inner sphere is $r_1 = 0.008\ m$

The radius of the outer sphere is $r _2 = 0.018 \ m$

The charge on the inner sphere is $q_1 = 3.62 *10^{-8} \ C$

The charge on the outer sphere is $q_2 = 1.62 *10^{-8} \ C$

The position from the origin is $d = 0.012 \ m$

Generally the electric field is mathematically represented as

$E = \frac{k (q_1 )}{ r^2}$

The reason for using $q_1$ for the calculation is due to the fact that the position considered is greater than the $r_1$ but less than $r_2$

Here k is the Coulomb constant with value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A{-2}$

So

$E = \frac{9*10^9 (3.62 *10^{-8}}{0.012^2}$

$E = 2.2625 *10^{6} \ N/C$

6 0

3 years ago

An object is moving with constant non-zero velocity. Which of the following statements about it must be true?

melisa1 [442]

Answer:

The net force on the object is zero.

Explanation:

An object is moving with constant non-zero velocity. If velocity is constant, it means that the change in velocity is equal to 0. As a result, acceleration of the object is equal to 0. Net force is the product of mass and acceleration. Hence, the correct option is (d) "The net force on the object is zero".

5 0

4 years ago