Ask question

Ask question

All categories

3 years ago

9

To practice Problem-Solving Strategy 16.2 Doppler effect. The sound source of a ship’s sonar system operates at a frequency of 2

2.0 kHz . The speed of sound in water (assumed to be at a uniform 20∘C) is 1482 m/s . What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling straight toward the ship at 4.95 m/s ? Assume that the ship is at rest in the water. mastering physics g

1 answer:

Whitepunk [10]3 years ago

7 0

Answer:

147 Hz

Explanation:

First we need to know what frequency will the waves hitting the whale have respect of it.

The change in frequency is calculated with this equation:

$\Delta f = \frac{\Delta v}{c} * f0$

However in this case, the wave is bouncing on the whale and coming back, so there is a double frequency shift (the difference is the same both times because the relative speeds between the two objects is the same both times)

$\Delta f = 2*\frac{\Delta v}{c} * f0$

Then:

$\Delta f = 2*\frac{\4.95 - 0}{1482} * 22000 = 147 Hz$

The frequency shift is positive, it will have increased frequency.

You might be interested in

A block of mass m1=3.7kg on a smooth inclined plane of angle 30 is connected by a cord over a small frictionless pulley to a sec

kari74 [83]

consider the forces on mass m₁ on the incline plane :

parallel to incline , force equation is given as

T - m₁ g Sin30 = m₁ a

T = m₁ g Sin30 + m₁ a eq-1

consider the force on mass m₂ on the incline plane :

m₂ g - T = m₂ a

T = m₂ g - m₂ a eq-2

Using eq-1 and eq-2

m₂ g - m₂ a = m₁ g Sin30 + m₁ a

inserting the values

(2.3 x 9.8) - 2.3 a = (3.7 x 9.8) Sin30 + 3.7 a

a = 0.74 m/s²

3 0

3 years ago

A VW beetle goes from zero to 27m/s in 7.6 seconds. What is the acceleration?

Firlakuza [10]

Initial speed(u)=0m/s
Final speed(v)= 27m/s
Time(t)=7.6s
Use the equation of motion: v = u + at
27 = 0 + a(7.6)
27/7.6 = a
a = 3.55 m/s^2 (3 s.f)

7 0

3 years ago

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the

Rama09 [41]

1) $1.11\cdot 10^{-7} J$

The capacitance of a parallel-plate capacitor is given by:

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

$r=\frac{2.0 cm}{2}=1.0 cm=0.01 m$

so the area is

$A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2$

While the separation between the plates is

$d=0.50 mm=5\cdot 10^{-4} m$

So the capacitance is

$C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F$

And now we can find the energy stored,which is given by:

$U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J$

2) 0.71 J/m^3

The magnitude of the electric field is given by

$E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m$

and the energy density of the electric field is given by

$u=\frac{1}{2}\epsilon_0 E^2$

and using

$E=4\cdot 10^5 V/m$ , we find

$u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3$

7 0

3 years ago

The concentration of an acid or base refers to how completely it dissociates in

Novosadov [1.4K]

Answer:

False

Explanation:

The Concentration of Acid or Base is the ph of the solution.

6 0

3 years ago

I will mark you as brainliest if you answer correctly

aleksklad [387]

(A) We can solve the problem by using Ohm's law, which states:
$V=IR$
where
V is the potential difference across the electrical device
I is the current through the device
R is its resistance
For the heater coil in the problem, we know $V=220 V$ and $R=220 \Omega$ , therefore we can rearrange Ohm's law to find the current through the device:
$I= \frac{V}{R}= \frac{220 V}{220 \Omega}=1 A$

(B) The resistance of a conductive wire depends on three factors. In fact, it is given by:
$R= \rho \frac{L}{A}$
where
$\rho$ is the resistivity of the material of the wire
L is the length of the wire
A is the cross-sectional area of the wire
Basically, we see that the longer the wire, the larger its resistance; and the larger the section of the wire, the smaller its resistance.

6 0

3 years ago