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3 years ago

Longitudinal waves Transverse waves

Physics

2 answers:

Natali [406]3 years ago

5 0

Answer:

longitudinal waves are waves that propagate in the same direction as the direction of movement of ìts particles, for example - sound

whereas

Transverse waves are waves that propagate perpendicular to the direction of movement of its particles, for example- light

galben [10]3 years ago

3 0

Answer:

In a transverse wave, the particles are displayed perpendicular to the direction the wave travels . in a longitudinal wave the particles are are displayed parallel to the direction the wave travels. an example of longitudinal waves in compressions moving along a slinky.

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A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns throu

Nady [450]

Answer:

The time interval is $t = 3 \ s$

Explanation:

From the question we are told that

The angular acceleration is $\alpha = 4.0 \ rad/s^2$

The time taken is $t = 4.0 \ s$

The angular displacement is $\theta = 80 \ radians$

The angular displacement can be represented by the second equation of motion as shown below

$\theta = w_i t + \frac{1}{2} \alpha t^2$

where $w_i$ is the initial velocity at the start of the 4 second interval

So substituting values

$80 = w_i * 4 + 0.5 * 4.0 * (4^2)$

=> $w_i = 12 \ rad/s$

Now considering this motion starting from the start point (that is rest ) we have

$w__{4.0 }} = w__{0}} + \alpha * t$

Where $w__{0}}$ is the angular velocity at rest which is zero and $w__{4}}$ is the angular velocity after 4.0 second which is calculated as 12 rad/s s

$12 = 0 + 4 t$

=> $t = 3 \ s$

4 0

3 years ago

Can you help me answer this?

Pavel [41]

The answer is D. If you aren't consistent with your drop positions, then your data may be invalid. To be frank: it basically screws over the experiment.

5 0

3 years ago

Consider the following statements about Newton's 2nd law in general. Select all of the statements that are true. Note, there may

Art [367]

Answer:

1. True

2. False

3. True

Explanation:

Newton's 2nd law states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:

$\sum F = ma$ (1)

where

$\sum F$ is the net force on the object

m is its mass

a is the acceleration

Furthermore, we know that acceleration is defined as the rate of change of velocity:

$a = \frac{dv}{dt}$

So let's now analyize the three statements:

1. A net force causes velocity to change: TRUE. Net force (means non-zero) causes a non-zero acceleration, which means that the velocity of the object must change.

2. If an object has a velocity, then we can conclude that there is a net force on the object: FALSE. The fact that the object has a velocity does not imply anything about its acceleration: in fact, if its velocity is constant, then its acceleration is zero, which would mean that the net force on the object is zero. So this statement is not necessarly true.

3. Accelerations are caused by the presence of a net force: TRUE. This is directly implied by eq.(1): the presence of the net force results in the object having a non-zero acceleration.

8 0

3 years ago

Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in

charle [14.2K]

Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

$t$ = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s

$s$ = distance to be moved by the rock long the horizontal = 98 yards

$y$ = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

$u$ = magnitude of initial velocity of the rock

$\theta$ = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

$\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\$

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

$\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\$

On dividing equation (1) by (2), we have

$\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ$

Now, putting this value in equation (2), we have

$u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s$

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

3 0

3 years ago

Which number is not found on the periodic table

miv72 [106K]

The letter “j” is never found on the periodic table. As for numbers, there’s an infinite amount

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3 years ago