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3 years ago

Acids are found in the pH range from 3 to 8, and bases are found from 10 to 17 on the pH scale. True False

1 answer:

6 0

False. Acids are actually found in the pH range from 0 to 7 and bases are found from 9 to 14. Meanwhile, 7 and 8 are considered neutral.

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All of the following describe the motion of an object except

Lostsunrise [7]

All describe an objects motion except : mass

8 0

3 years ago

An ambulance is rapidly approaching you at a stop light. What happens to the frequency and pitch of the sound as the ambulance d

vagabundo [1.1K]

Answer:

A )

Explanation:

This change in frequency observation occur due to doppler effect

if the wave source moves,In the time between one wave peak being emitted and the next, the source will have moved so that the shells will no longer be concentric. The wavefronts will get closer together in front of the source as it travels and will be further apart behind it. (see the graph)

when the person standing still in front of the ambulance, he will observe a <em>higher frequency </em>than before as the source travels towards them.

$f_(observed)=\frac{V_(wave)}{V_(wave)-V_(source)} *f_(original)$

The pitch we hear depends on the frequency of the sound wave.

A high frequency corresponds to a high pitch

as we hear a higher frequency , it makes the <em>pitch higher</em> too

5 0

3 years ago

Read 2 more answers

A skateboarder is moving at 1.75 m/s when she starts going up an incline that causes an acceleration of -0.20 m/s2

Rudiy27

Answer:

Approximately $7.66\; \rm m$ .

Explanation:

<h3>Solve this question with a speed-time plot</h3>

The skateboarder started with an initial speed of $u = 1.75\; \rm m \cdot s^{-1}$ and came to a stop when her speed became $v = 0\; \rm m \cdot s^{-1}$ . How much time would that take if her acceleration is $a = -0.20\; \rm m \cdot s^{-1}$ ?

$\begin{aligned} t &= \frac{v - u}{a} \\ &= \frac{0\; \rm m \cdot s^{-1} - 1.75\; \rm m \cdot s^{-1}}{-0.20\; \rm m \cdot s^{-2}} \approx 8.75\; \rm s\end{aligned}$ .

Refer to the speed-time graph in the diagram attached. This diagram shows the velocity-time plot of this skateboarder between the time she reached the incline and the time when she came to a stop. This plot, along with the vertical speed axis and the horizontal time axis, form a triangle. The area of this triangle should be equal to the distance that the skateboarder travelled while she was moving up this incline until she came to a stop. For this particular question, that area is approximately equal to:

$\displaystyle \frac{1}{2} \times 1.75\; \rm m \cdot s^{-1} \times 8.75\; \rm s \approx 7.66\; \rm m$ .

In other words, the skateboarder travelled $15.3\; \rm m$ up the slope until she came to a stop.

<h3>Solve this question with an SUVAT equation</h3>

A more general equation for this kind of motion is:

$\displaystyle x = \frac{1}{2}\, (u + v) \, t = \frac{1}{2}\, (u + v)\cdot \frac{v - u}{a}= \frac{v^2 - u^2}{2\, a}$ ,

where:

$u$ and $v$ are the initial and final velocity of the object,
$a$ is the constant acceleration that changed the velocity of this object from $u$ to $v$ , and
$x$ is the distance that this object travelled while its velocity changed from $u$ to $v$ .

For the skateboarder in this question:

$\begin{aligned}x &= \frac{v^2 - u^2}{2\, a}\\ &= \frac{\left(0\; \rm m \cdot s^{-1}\right)^2 - \left(1.75\; \rm m \cdot s^{-1}\right)^2}{2\times \left(-0.20\; \rm m \cdot s^{-2}\right)}\approx 7.66\; \rm m \end{aligned}$ .

6 0

4 years ago

Hiran is standing beside the road when he hears a bird flying away from hip and chirping. The bird’s chirp has a frequency of 18

Murrr4er [49]

The frequency of bird chirping hear by hiran will be 1.77 kHz.

<u>Explanation:</u>

As per Doppler effect, the observer will feel a decrease in the frequency of the receiving signal if the source is moving away from the observer. So the shifted frequency is obtained using the below equation:

$f'=\frac{c}{c+v_{s} }f$

Here , c is the speed of sound, Vs is the velocity of source with which it is moving away. f is the original frequency of source and f' is the frequency shift heard by the observer.

As here, f = 1800 Hz, Vs= 6 m/s and c = 343 m/s, then

$f'=\frac{343}{343+6} \times 1800=1.77\ kHz$

So, the frequency of bird chirping hear by hiran will be 1.77 kHz.

4 0

3 years ago

Lol somebody tell me the answer ill also give 5 stars

AnnyKZ [126]

Answer:

i believe it's carbon. i may be wrong. but i haven't done atoms in almost 2 years. sorry if it's wrong. :)) hava good day

Explanation:

4 0

3 years ago

Read 2 more answers