All categories

2 years ago

The speed of sound in seawater is 1470 m/s. A dolphin sends out a click that reflects off of an

Physics

1 answer:

Nitella [24]2 years ago

3 0

Answer: 0.204 s

Explanation:

The speed of sound $V$ is defined as the distance traveled $d$ in a especific time $t$ :

$V=\frac{d}{t}$

Where:

$V=1470 m/s$ is the speed of sound in seawater

$t$ is the time the sound wave travels from the dolphin and then returns after the reflection

$d=2(150 m)$ is twice the distance between the dolphin and the object to which the sound waves are reflected

Finding $t$ :

$t=\frac{d}{V}$

$t=\frac{2(150 m)}{1470 m/s}$

<u>Finally:</u>

$t=0.204 s$

You might be interested in

Convert the volume 8.06 in.3 to m3, recalling that1in. =2.54cmand100cm=1m. Answer in units of m3.

galina1969 [7]

1 in=2.54 cm=(2.54 cm)(1 m/100 cm)=0.0254 m
Therefore:
1 in=0.0254 m
1 in³=(0.0254 m)³=1.6387064 x 10⁻⁵ m³

Therefore:

8.06 in³=(8.06 in³)(1.6387064 x 10⁻⁵ m³ / 1 in³)≈1.321 x 10⁻⁴ m³.

Answer: 8.06 in³=1.321 x 10⁻⁴ m³

8 0

2 years ago

Your friend says, “chemical changes are caused by an input in energy. In physical changes, there is no transfer of energy” is yo

nalin [4]

Answer:

Ok, let's suppose the simplest of the physical changes:

We have an object that is not moving (so it is not accelerated)

and there is change, now the object moves.

Because there was a change, means that there was an acceleration, and by the second Newton's law.

Force equals mass times acceleration:

F = m*a

There must be a force.

So suppose that you pushed the object, then some energy that you had, you transferred it to the object, that now is moving and now has kinetic energy.

Now, is kinda true that in a closed system the total energy is always constant, but it depends on what is our system.

So if we think in our system as you and the object, then in the whole system the energy does not change because the energy that you lost is now on the object, but again, there was a transfer of energy.

So no, your friend is not correct.

3 0

2 years ago

A force Fof 40000 lbf is applied to rod AC the negative Y-direction. The rod is 1000 inches tall. A Force P of 25 lbf is applied

erastova [34]

Answer:

The answer is "effective stress at point B is 7382 ksi "

Explanation:

Calculating the value of Compressive Axial Stress:

$\to \sigma y =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\$

Calculating Shear Transverse:

$\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}$

$= \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi$

$= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\ in$

$\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\$

$= [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi$

8 0

2 years ago

During a race the dirt bike was observed to leap up off the small hill at A at an angle of 60^o with the horizontal. If the poin

const2013 [10]

Answer:

Velocity is equal to $27.3$ feet per second and time is equal to $1.4668$ seconds

Explanation:

Given

The horizontal distance traveled by dirt bike before landing $= 20$ feet

Angle of flight $= 60$ degree

As we know that Horizontal distance (H) is equal to

$= H_0 + V_0 * t\\$

Where $H_0$ is the initial horizontal distance

$V_0$ is the velocity with which the bike is travelling in horizontal direction

and $t$ is the time in seconds

Substituting the given values, we get -

$H = H_0 + v*t\\20 = 0 + v * cos \theta * t\\20 = v * cos 60 * t\\t = \frac{20}{v * cos 60} \\t = \frac{40}{v}$

Now distance traveled in vertical direction is equal to

$Y = Y_0 + v_0 * t + \frac{1}{2} a * t^2$

here acceleration will be equal to acceleration due to gravity which is equal to $- 32.2 \frac{ft}{s^2}$ . It is negative as its is acting in upward direction

Thus,

$Y = 0 + v * sin 60 + \frac{1}{2} * (-32.2) * (\frac{40}{v} )^2\\0 = 0 + 0.866 v + \frac{-25760}{v^2} \\0.866 v = \frac{-25760}{v^2}\\v^3 = \frac{-25760}{0.866} \\v = 27.3$

Velocity is equal to $27.3$ feet per second and time is equal to $1.4668$ seconds

8 0

2 years ago

The world record for pole vaulting is 6.15 m. If the pole vaulter's gravitational potential energy is 4942 J, what is his mass?

navik [9.2K]

The gravitational potential energy is calculated by multiplying the mass of the object to the height and the gravitational acceleration which is 9.8 m/s^2. We do as follows:

GPE = mgh
GPE = 4942
4942 = m (9.8)(6.15)
m = 82 kg

Hope this helps.

7 0

2 years ago