Please answer the question on the picture

A stationary sub uses sonar to send a 1.18x10^3 hertz sound wave through ocean water. The reflected sound wave from the flat oce

vampirchik [111]

Answer:

a) v = 1524.7 m/s

b) T = 8.47*10^-4 s

λ = 1.29 m

Explanation:

a) First, in order to calculate the speed of the sound wave, you take into account that the velocity is constant, then, you use the following formula:

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

d: distance traveled by the sound wave, which is twice the distance to the ocean bottom = 2*324 m = 648 m

t: time that sound wave takes to return to the sub = 0.425

$v=\frac{648m}{0.425s}=1524.7\frac{m}{s}$

hence, the speed of the sound wave is 1524.7 m/s

b) Next, with the value of the velocity of the wave you can calculate the wavelength of the wave, by using the following formula:

$v=\lambda f\\\\\lambda=\frac{v}{f}$

f: frequency = 1.18*10^3 Hz

$\lambda=\frac{1524.7m/s}{1.18*10^3s^{-1}}=1.29m$

And the period is:

$T=\frac{1}{f}=\frac{1}{1.18*10^3s^{-1}}=8.47*10^{-4}s$

hence, the wavelength and period of the sound wave is, respectively, 1.29m and 8.47*10^-4 s

5 0

3 years ago

When charges qa, qb, and qc are placed respectively at the corners a, b, and c of a right triangle, the potential at the midpoin

earnstyle [38]

Answer:

8v

Explanation:

First we apply super position principle

Vt= v1 + v2+ v3

Remove qa

But vt= 20v

So V = v2+v3

V1= 20-15

= 5v

Remove qb

V= v1+v3

V=8v

So the potential when qa and qc are remove is the potential due to qb

Which is 8v

7 0

3 years ago

A car travels 60 km in the first 2 hours and 68 km in the next 2 hours.

ZanzabumX [31]

Answer:32km/hr

Explanation:

Speed=distance/time

Speed=60/2 =30km/hr speed=68/2=34km/he

Average speed=(30+34)/2 =64/2=32km/hr

4 0

3 years ago

Why will a sheet of paper fall slower than one that is crumbled into a ball

Zarrin [17]

It’s coming in contact with more air molecules than I would if it was in a ball because there is less surface area

4 0

3 years ago

Read 2 more answers

What is the total energy q released in a single fusion reaction event for the equation given in the problem introduction? use c2

kkurt [141]

Answer:

$4.3\cdot 10^{-12} J$

Explanation:

The fusion reaction in this problem is

$4^1_1H \rightarrow ^4_2He +2e^+$

The total energy released in the fusion reaction is given by

$\Delta E = c^2 \Delta m$

where

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\Delta m$ is the mass defect, which is the mass difference between the mass of the reactants and the mass of the products

For this fusion reaction we have:

$m(^1_1H)=1.007825u$ is the mass of one nucleus of hydrogen

$m(^4_2 He)=4.002603u$ is the mass of one nucleus of helium

So the mass defect is:

$\Delta m =4m(^1_1 H)-m(^4_2 He)=4(1.007825u)-4.002603u=0.028697u$

The conversion factor between atomic mass units and kilograms is

$1u=1.66054\cdot 10^{-27}kg$

So the mass defect is

$\Delta m =(0.028697)(1.66054\cdot 10^{-27})=4.765\cdot 10^{-29}kg$

And so, the energy released is:

$\Delta E=(3.0\cdot 10^8)^2(4.765\cdot 10^{-29})=4.3\cdot 10^{-12} J$

6 0

3 years ago