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2 years ago

Sound

2 answers:

4 0

It is a longitudinal wave. In a longitudinal wave the particles of the medium vibrate in a direction parallel to the direction of energy transport. Sound waves are mechanical waves as well as longitudinal waves that require a medium. They do not travel through a vacuum.

Ann [662]2 years ago

3 0

Longitudinal, because the sound can only travel at one direction

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If the force vector has a magnitude of F = 12.5 N, the displacement has a length of Δr = 0.75 m, and the angle between the vecto

Thepotemich [5.8K]

Answer: 8.81 J

Explanation: The scalar or dot product of 2 vector equals the product of the magnitude of both vectors and the cosine value of the angle between them

Magnitude of vector f (force) as given in the question = 12.5N

Magnitude of length = 0.75m

Angle between vectors = 25°

F. Δr = 12.5 * 0.75 * cos 25

F. Δr = 12.5 * 0.75 * 0.9063

F. Δr = 8.81 J

7 0

2 years ago

An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the

garik1379 [7]

(a) 6.04 rev/s

The speed of the ball is given by:

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the ball from the centre of the circle

In situation 1), we have

$\omega=8.13 rev/s \cdot 2\pi = 51.0 rad/s$

r = 0.600 m

So the speed of the ball is

$v=(51.0 rad/s)(0.600 m)=30.6 m/s$

In situation 2), we have

$\omega=6.04 rev/s \cdot 2\pi = 37.9 rad/s$

r = 0.900 m

So the speed of the ball is

$v=(37.9 rad/s)(0.900 m)=34.1 m/s$

So, the ball has greater speed when rotating at 6.04 rev/s.

(b) $1561 m/s^2$

The centripetal acceleration of the ball is given by

$a=\frac{v^2}{r}$

where

v is the speed

r is the distance of the ball from the centre of the trajectory

For situation 1),

v = 30.6 m/s

r = 0.600 m

So the centripetal acceleration is

$a=\frac{(30.6 m/s)^2}{0.600 m}=1561 m/s^2$

(c) $1292 m/s^2$

For situation 2 we have

v = 34.1 m/s

r = 0.900 m

So the centripetal acceleration is

$a=\frac{v^2}{r}=\frac{(34.1 m/s)^2}{0.900 m}=1292 m/s^2$

5 0

2 years ago

A sonar emits a sound signal of frequency 40000Hz, towards the bottom of the sea.

kap26 [50]

Answer:

0.000025s

Explanation:

Period it’s. : T(s)= 1/f(Hz)=1/40000Hz=0.000025s

8 0

2 years ago

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

8 0

2 years ago

Read 2 more answers

Which one do I press guys?

taurus [48]

Answer:

The Nucleus...

Explanation:

8 0

2 years ago