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3 years ago

Dolphin echolocation is similar to ultrasound. Reflected sound waves

Physics

1 answer:

s344n2d4d5 [400]3 years ago

3 0

Answer:

Waves with high frequencies have shorter wavelengths that work better than low frequency waves for successful echolocation.

Explanation:

To understand why high-frequency waves work better than low frequency waves for successful echolocation, first we have to understand the relation between frequency and wavelength.

The relation between frequency and wavelength is given by

λ = c/f

Where λ is wavelength, c is the speed of light and f is the frequency.

Since the speed of light is constant, the wavelength and frequency are inversely related.

So that means high frequency waves have shorter wavelengths, which is the very reason for the successful echolocation because waves having shorter wavelength are more likely to reach and hit the target and then reflect back to the dolphin to form an image of the object.

Thus, waves with high frequencies have shorter wavelengths that work better than low frequency waves for successful echolocation.

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On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on t

kozerog [31]

Answer:

a) 6 times farther. b) 6 times longer.

Explanation:

Once released, in the horizontal direction, no other forces act on the ball, so it continues moving at the same initial velocity, which is given by the projection of the velocity vector in the horizontal direction, as follows:

vₓ = v* cos (25º) = 23 m/s * 0.906 = 20.8 m/s

In the vertical direction, the initial velocity is the projection of the velocity vector along the vertical axis, as follows:

vy = v* sin (25º) = 23 m/s * 0.422 = 9.72 m/s

Assuming that the acceleration is constant, and equal to 1/6*g, we can calculate the total time of flight, with the following kinematic equation for the vertical displacement:

$y = voy*t - (\frac{1}{2}*\frac{g}{6} * t^{2} )$

If the total displacement in the vertical direction is 0 (which means that the time if the total time of flight), we can solve for t, as follows:

$t = \frac{voy*12}{g} = \frac{9.72 m/s*12}{9.8m/s2} = 11. 9 s$

On earth, this time could be calculated in the same way:

$t = \frac{voy*12}{g} = \frac{9.72 m/s*2}{9.8m/s2} = 1.98 s$

As the time is defined by the vertical movement, we can find the horizontal distance travelled on the moon, as follows:

Δx = v₀ₓ * t = 20.8 m/s * 11. 9 s = 248.1 m

On earth, the distance travelled had been as follows:

Δx = v₀ₓ * t = 20.8 m/s * 1.98 s = 41.3 m

⇒ Δx(moon) / Δx(earth) = 248.1 / 41.3 = 6.00

b) As we have just found, the time of flight on the moon and on the earth are as follows:

tmoon = 11. 9 s

tearth = 1.98 s

⇒ t(moon) / t(earth) = 11.9 / 1.98 = 6.0

8 0

2 years ago

A ball rolls with a speed of 2.0m/s across a level table that is 1.0m above the floor. Upon reaching the edge of the table it fo

Mekhanik [1.2K]

Answer:0.58 m

Explanation:

The initial velocity of the ball is u = 2.0 m/s

The height of the table is, h = 1.0 m

The ball falls in vertical direction under acceleration due to gravity.

Time taken for ball to hit the floor:

h= ut + 0.5gt² ( from the equation of motion)

1.0 m=2.0 m/s × t+0.5 × 9.8 m/s²× t²

Solving this for t,

t = 0.29 s ( we have neglected the negative value of t)

In the same time, the ball would cover a horizontal distance of :

s = u t

⇒s = 2.0 m/s×0.29 s = 0.58 m

Thus, the landing spot is 0.58 m away from the table.

6 0

3 years ago

Read 2 more answers

How do you find the force of a ping-pong ball rolling down a track?

Kaylis [27]

Answer:

Weight

a) weight's vertical component = Normal upward force

b) weight's horizontal component = Friction force = (mass of ball)(acceleration)

These forces depend upon the track,

1) inclined or horizontal

2) steepness.

Explanation

The force of gravity points straight down, but a ball rolling down a ramp doesn't go straight down, it follows the ramp. Therefore, only the component of the weight which points along the direction of the ball's motion can accelerate the ball.

weight's horizontal component = Friction force = (mass of ball)(acceleration)

The other component pushes the ball into the ramp, and the ramp pushes back.

If the ramp is horizontal, then the ball does not accelerate, as gravity pushes the ball into the ramp and not along the surface of the ramp. Hope this helps. Can u give me brainliest

Explanation:

3 0

2 years ago

A cylinder of gas is at room temperature (20°C). The air conditioner breaks down, and the temperature rises

noname [10]

Answer: we divide this equation by pV and use {C}_{p}={C}_{V}+ . ... The temperature, pressure, and volume of the resulting gas-air mixture

Explanation:

4 0

3 years ago

Lagrangian mechanics. Determine the equations of motion for a particle of mass m constrained to move on the surface of a cone in

maria [59]

Answer:

Explanation:

Hi!

In order to obtain the Lagrangian of the system we must first write the Kinetic and Potential Energies. Lets orient our axes such that the axis of the cone coincide with the z axis. In cilindrical coordinates we have

$v^{2} = \frac{dr}{dt}^{2} +r^{2} \frac{d\theta }{dt} ^{2} +\frac{dz}{dt} ^{2}$ - (1)

But, since the particle is constrained to move on the surface of the cilinder, we have the following relation between r and z:

$\frac{r}{z}=tan(45)$

or:

$z = r cot(45)$ - (2)

and:

$\frac{dz}{dt} = \frac{dr}{dt} cot(45)$

replacing (2) in (1) we obtain:

$v^{2} = \frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}$ - (3)

Now the kinetic energy is given as:

$T = \frac{1}{2}m(\frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2})$ - (4)

And the potential energy is given by:

$V = -mgz = -mgr cot(45)$

So the Langrangian is given by:

$L = T - V= \frac{1}{2}m(\frac{dr}{dt}^{2}(1+cot(45)+r^{2})\frac{d\theta }{dt} ^{2}) + mgr cot(45)$

And the equations of motion are:

For θ

$\frac{d}{dt} (mr\frac{d\theta}{dt}) = 0-->mr{d\theta}{dt}=c$

For r

$\frac{d}{dt}(m\frac{dr}{dt}(1+cot(45) )= mgcot(45)+mr\frac{d\theta}{dt} ^{2}\\m\frac{d^{2} r}{dt^{2} }(1+cot(45)= mgcot(45)+mr\frac{d\theta}{dt} ^{2}$

Obtained from the Euler-Langrange equations

Here the conserved quantity is given by the first equation of motion, namely:

$mr\frac{d\theta}{dt}=c$

Which is the magnitude of the angular momentum

7 0

3 years ago