Ask question

Ask question

All categories

1 year ago

13

As a wave travels through a medium, the particles of the medium vibrate in a direction perpendicular to the direction of the wav

e’s motion. What kind of wave is this? Longitudinal Wave, Pulse Wave, Transverse Wave, Pressure Wave, or Electromagnetic Wave?

2 answers:

Readme [11.4K]1 year ago

7 0

The answer is Transverse wave

Cloud [144]1 year ago

7 0

The answer is transverse wave

You might be interested in

A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the

ser-zykov [4K]

Answer:

The maximum length during the motion is $L_{max} = 1.45m$

Explanation:

From the question we are told that

The mass is $m =0.5 kg$

The vertical spring length is $L = 1.10m$

The unstretched length is $L_{un} = 1.30m$

The initial speed is $v_i = 1.3m/s$

The new length of the spring $L_{new} = 1.30 m$

The spring constant k is mathematically represented as

$k = -\frac{F}{y}$

Where F is the force applied $= m * g = 0.5 * 9.8=4.9N$

y is the difference in weight which is $=1.10-0.50=0.6m$

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

Now substituting values accordingly

$k = \frac{4.9}{0.6}$

$= 8.17 N/m$

The elastic potential energy is given as $E_{PE} = \frac{1}{2} k D^2$

where D is this the is the displacement

Since Energy is conserved the total elastic potential energy would be

$E_T = initial \ elastic\ potential \ energy + kinetic \ energy$

$E_T = \frac{1}{2} k D_{max}^2 = \frac{1}{2} k D^2 + \frac{1}{2} mv^2$

Substituting value accordingly

$\frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2$

$4.085 * D_{max}^2 = 3.69$

$D^2_{max} = 0.9033$

$D_{max} = 0.950m$

So to obtain total length we would add the unstretched length

So we have

$L_{max} = 0.950 + 0.5 = 1.45m$

5 0

2 years ago

Read 2 more answers

Give the relationship(s) for any pair of protons with the proper term(s). Label – your choice. A.Heterotopic B.Heterotopic, dias

Afina-wow [57]

Answer and Explanation

• Heterotopic protons are those that when substituted by the same substituent, are structurally different. They are not similar, diastereotopic or enantiotopic.

• Diastreotopic protons refers to two protons in a molecule which, if replaced by the same substituent, would generate compounds that are diastereomers. Diastereotopic groups are often, but not always, identical groups attached to the same atom in a molecule containing at least one chiral center.

For example, the two hydrogen atoms of the C3 carbon in (S)-2-bromobutane are diastereotopic (shown in the attached image). Replacement of one hydrogen atom with a bromine atom will produce (2S,3R)-2,3-dibromobutane. Replacement of the other hydrogen atom with a bromine atom will produce the diastereomer (2S,3S)-2,3-dibromobutane.

• Homotopic protons in a compound are equivalent protons. Two protons A and B are homotopic if the molecule remains the same (including stereochemically) when the protons are interchanged with some other atom (substituent) while the remaining parts of the molecule stay fixed. Homotopic atoms are always identical, in any environment.

For example, ethane, the two H atoms on C1 and C2 carbons on the same side (as shown in the attached image) are homotopic as they exhibit the phenomenon described above.

• Enantiotopic protons are two protons in a molecule which, if one or the other were replaced (by the same substituent), would generate a chiral compound. The two possible compounds resulting from that replacement would be enantiomers.

For example, in the attached image to this answer, the two hydrogen atoms attached to the second carbon in butane are enantiotopic. Replacement of one hydrogen atom with a bromine atom will produce (R)-2-bromobutane. Replacement of the other hydrogen atom with a bromine atom will produce the enantiomer (S)-2-bromobutane.

Hope this helps!!!

7 0

2 years ago

Oduvanchick [21]

Answer:

$-22.7 m/s^2$

Explanation:

This is a uniformly accelerated motion, so we can determine the deceleration of the car by using a suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the car in this problem,

u = 27.8 m/s

v = 0

s = 17 m

Solving for a, we find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{0-27.8^2}{2(17)}=-22.7 m/s^2$

4 0

3 years ago

A brick is thrown from a wall that is 19.2 m above the ground and hits the ground 5.00 s later. What was the initial velocity of

Brums [2.3K]

Answer:

0

Explanation:

Given the following :

Height of wall = 19.2 m

Time taken to hit the ground = 5 seconds

Acceleration due to gravity (g) = 9.8m/s ( downward motion)

The initial velocity of the object refers to the Velocity of the object at time t = 0

Initial Velocity = g × time

Initial Velocity = 9.8 × 0 = 0

5 0

3 years ago

Two forces act on a 41-kg object. One force has magnitude 65 N directed 59° clockwise from the positive x-axis, and the other ha

Genrish500 [490]

Answer:

a = 1.41 m/s²

Explanation:

Given that

mass ,m= 41 kg

F₁ = 65 N , θ = 59°

F₂ = 35 N ,θ = 32°

The component of Force F₁

F₁x= F₁cos59° i

F₁x= 65 x cos59° i = 33.47 i

F₁y= - F₁ sin 59° j

F₁y= - 65 x sin 59° j = - 55.71 j

The component of Force F₂

F₂x= F₂ sin 32° i

F₂x= 35 x sin 32° i = 18.54 i

F₂y= F₂ cos 32° j

F₂y= 35 x cos 32° j = 29.68 j

The total force F

F= 33.47 i + 18.54 i - 55.71 j + 29.68 j

F= 52.01 i - 26.03 j

The magnitude of the force F

$F=\sqrt{52.01^2+26.03 ^2}\ N$

F=58.16 N

We know that

F= m a

a= Acceleration

m=mass

58.16 = 41 x a

a = 1.41 m/s²

6 0

3 years ago