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maks197457 [2]
1 year ago
10

The speed of sound in water is 1,492 m/s. A sonar signal is sent straight down from a ship at a point just below the water's sur

face and 7.5 s later the reflected signal is detected. How deep is the ocean beneath the ship?
Physics
1 answer:
ASHA 777 [7]1 year ago
6 0

Answer:

5595 m

Explanation:

The wave travels down   (the depth) and the back up (the depth again)

  this distance is     2 d

2 d =  1492 m/s * 7.5 s

2d = 11190 m

d = 5595 m

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A charge moves a distance of 1.8 cm in the direction of a uniform electric field having a magnitude of 214 N/C. The electrical p
Andrei [34K]

Answer:

 13.4 x 10 raise to power -19  C

Explanation:

. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m

. The uniform electric field is  E = 214 N/M

, The decrease in electrical potential energy is   d(P.E) = 51.63 x 10 raise to power -19 J

Let the magnitude of the charge of the moving particle be q

which is given by the equation

d(P.E) =qEd

51.63 x 10 power -19 = q(214)(0.018)

51.63 x 10 power -19 =3.852q

by making q the formular,

q = 13.4 x 10 power -19 C  

5 0
3 years ago
Eukaryotic cells can be unicellular and multicellular true or false
dimaraw [331]
Yes, it can be unicellular and multicellular
8 0
3 years ago
Which is the resulting direction of a surface wave
Sliva [168]
In a surface wave (like water) the water goes up and down, but the wave travels across (parallel to) the surface.
3 0
2 years ago
he calculation of kinetic energy uses the formula, KE = 1 2 mv2. Identify the unit that is required for mass and velocity, respe
Umnica [9.8K]
Mass is measured in kg
Velocity is measured in ms^-1
Hope this is what you were looking for
5 0
2 years ago
A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.
Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity = 1.5 ra/s

\alpha = Angular acceleration

\theta = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2

Acceleration while slowing down is -0.0047 rad/s²

t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s

Time taken to slow down is 335.103 seconds

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0

Solving the equation

t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0
2 years ago
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