Ask question

Ask question

All categories

2 years ago

9

Standing at a crosswalk, you hear a frequency of 550 Hz from the siren of an approaching ambulance. After the ambulance passes,

the observed frequency of the siren is 475 Hz. Determine the ambulance's speed from these observations. (Take the speed of sound to be 343 m/s.)

1 answer:

FromTheMoon [43]2 years ago

6 0

There are six steps to this process , I uploaded step one and as you can see you can get all six on Quizlet:). Good luck

You might be interested in

A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o

Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let $q_1$ and $q_2$ be the charges such that

$q_1+q_2=4.7$

and Force between charge particles is given by

$F=\frac{kq_1q_2}{r^2}$

$4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}$

$q_1\cdot q_2=0.522$

put the value of $q_1$

$q_2\left ( 4.7-q_2\right )=0.522$

$q_2^2-4.7q_2+0.522=0$

$q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}$

$q_2=0.114 C$

thus $q_1=4.586 C$

3 0

2 years ago

How does the balloon react to the cloth item?

almond37 [142]

Answer: When rubbing a balloon with a wool cloth, it puts negative charges on the balloon. Negative charges attract to positive charges. If a balloon is not rubbed with the wool cloth, it has an equal amount of negative to positive charges, so it will attract to a rubbed balloon.

8 0

2 years ago

Another term for technology is

Katyanochek1 [597]

Answer:

a. applied science

Explanation:

$.$

3 0

2 years ago

Read 2 more answers

What is the specific heat of an object if it takes 1200 J to raise the temperature of a 20 kg object by 6 degrees C?

ch4aika [34]

Answer:

$c=10\ J/kg^{\circ} C$

Explanation:

Given that,

Heat required, Q = 1200 J

Mass of the object, m = 20 kg

The increase in temperature, $\Delta T=6^{\circ} C$

We need to find the specific heat of the object. The heat required to raise the temperature is given by :

$Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{1200}{20\times 6}\\\\c=10\ J/kg^{\circ} C$

So, the specific heat of the object is $10\ J/kg^{\circ} C$ .

5 0

2 years ago

#1 Not sure where to start. This is for AP Physics!

yaroslaw [1]

First,

$\rho=\dfrac mV$

where $\rho$ is density, $m$ is mass, and $V$ is volume. We can compute the volume of the roll:

$2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V$

$\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3$

When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness $x$ . Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

So we have

$V=Ax$

where $A$ is the given area, so

$4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x$

$\implies x\approx0.255\,\mathrm m=255\,\mathrm{mm}$

If we're taking significant digits into account, the volume we found would have been $V=470\,\mathrm m^3$ , in turn making the thickness $x=250\,\mathrm{mm}$ .

8 0

2 years ago