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Alexxandr [17]
3 years ago
7

Was the object accelerating during the time interval labeled “B”? A) Yes B) No

Physics
2 answers:
Oxana [17]3 years ago
7 0
Yes it was accelerating
Alla [95]3 years ago
5 0

There's not enough information in the graph to determine whether it was or it wasn't.  It could have been, but we can't tell.

Before I explain this answer, it's VERY important to know exactly what "accelerating" means:

-- "Accelerating" does NOT always mean "speeding up".

-- "Accelerating" means ANY change in velocity.

-- That can be ==> <u>speeding up</u>, or ==> <u>slowing down</u>, or ==> <u>turning</u>.

This graph is labeled "Velocity vs Time", but it isn't really. "Velocity" means <u>speed</u> AND <u>direction</u>. The graph only shows us the object's speed, but nothing about its direction.

During time-interval-B, the speed isn't changing, but the object could still be changing <u>direction</u>, like turning a corner, or on a curved path. We don't know. The graph doesn't show us that information.

The object could be accelerating during time-interval-B. It's true that the speed wasn't changing. But maybe the direction was. If the direction was changing, then we have to say the object was 'accelerating' during the interval labeled B .  

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Sound waves are also called compression waves. This means that as the wave travels through air, the ____ in creases and decrease
Alja [10]
The speed creases and decreases
5 0
3 years ago
You drive 6.00 km at 50.0 km/h and then another 6.00kmat 900 km/h Your average speed over
Cerrena [4.2K]

Explanation:

average speed = total distance travelled / total time travelled

time to travel the first 6km: 6 / 50 = 3/25 (h)

time to travel the next 6km: 6 / 90 = 1/15 (h)

[I think there's problem in the question 'cause 900km/h sounds impossible for normal person to travel in normal condition]

The total time: 3/25 + 1/15 = 14/75 (h)

Average speed over the 12 km drive will be:

\frac{12}{ \frac{14}{75} }  =  \frac{450}{7}  = 64.3 \: km{h}^{ - 1}

8 0
3 years ago
The amplitude of a standing sound wave in a long pipe closed at the left end is sketched below.
ollegr [7]

(1) The harmonic number for the mode of oscillation is 3.

(2) The pitch (frequency) of the sound is 579.55 Hz

(3) The level of the water inside the vertical pipe is 0.1 m.

<h3>The harmonic number</h3>

The harmonic number for the mode of oscillation illustrated for the closed pipe is 3.

<h3>Frequency of the wave</h3>

The pitch (frequency) of the sound is calculated from third harmonic formula;

f = 3v/4L

where;

  • v is speed of sound
  • L is length of the pipe

f = (3 x 340) / (4 x 0.44)

f = 579.55 Hz

<h3>level of the water</h3>

wave equation for first harmonic of a closed pipe is given as

f  = v/(4L)

251.1  = 340/(4L)

4L = 340/251.1

4L = 1.35

L = 1.35/4

L = 0.34 m

level of water = 0.44 m - 0.34 m = 0.1 m

Thus, the level of the water inside the vertical pipe is 0.1 m.

Learn more about harmonics of closed pipes here: brainly.com/question/27248821

#SPJ1

3 0
1 year ago
An object that completes 20 vibrations in 10 seconds has a frequency of
nika2105 [10]

Answer:

<em> The object has frequency of 2 Hz and time period of 0.5 s.</em>

Explanation:

<em>Frequency</em> is defined as number of oscillation per second ie back and forth swings done in single second.

Here it is given that the object oscillates 20 times in 10 seconds.

So f = \frac{20}{10} = 2Hz

The <em>time period</em> is defined as time taken by the object to complete one full oscillation.

T = \frac{1}{f}

T= \frac{1}{2} =0.5 s

<em>Thus the object has frequency of 2 Hz and time period of 0.5 s.</em>

7 0
3 years ago
A person drives a car around a circular cloverleaf with a radius of 77 m at a uniform speed of 10 m/s. What is the acceleration
umka2103 [35]

Answer:

770m/s

Explanation:

caculation using one of the newton law of motion

6 0
3 years ago
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