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Mandarinka [93]
3 years ago
13

The speed of a wave is 1500 m/s. What is the frequency of the wave if the wavelength is 2 m?

Physics
1 answer:
Sonja [21]3 years ago
5 0

Answer:

Explanation:

1500 m/s / 2 m = 750 Hz

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Which of the following is correct for speed?
Olin [163]

Answer:

Speed is a "scalar" quantity

(C) is the correct answer

An object could travel at 10 m/s to some point and then return to the origin at 10 m/s for an average speed of 10 m/s, however it's displacement over that time would be zero for a net velocity of zero.

4 0
3 years ago
For a wave, what term is defined as the maximum height of a crest, or depth of a trough, relative to the normal level? for a wav
sammy [17]
The appropriate response is amplitude. Amplitude is the target estimation of the level of progress (positive or negative) in environmental weight (the pressure and rarefaction of air atoms) created by sound waves.

I hope the answer will help you. 
6 0
3 years ago
Tony drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took hours. When Tony drove hom
polet [3.4K]

Answer:

Question not completed, so I analysed the question first

Tony drove to the mountains last weekend. there was heavy traffic on the way there, and the trip took 6 hours. when tony drove home, there was no traffic and the trip only took 4 hours. if his average rate was 22 miles per hour faster on the trip home, how far away does tony live from the mountains?

Explanation:

Let use variables to solve the problems

Let the first trip to be mountain take x hours

Let the trip back home take y hours

Let the speed to while going to the mountain be a miles/hour

Then, while going home it was b miles/hour faster than while going to the mountain.

Then, speed going home is (a+b)miles / hour

The formula for speed is given as

Speed=distance/time

The constant through out the journey is distance, the two journey has the same distance.

Then,

Distance =speed×time

For first journey going to the mountain

Distance = a×x=ax miles

For the second journey going home

Distance =y×(a+b)

Distance Mountain= distance home

ax=y(a+b)

Make a subject of the formula

ax=ya+yb

ax-ya=yb

a(x-y)=yb

a=yb/(x-y)

Therefore, distance from mountain is

Distance=speed ×time

Distance= a×x=ax

Now, applying the questions

So from the questions

x=6hours, y=4hours

Also, b=22miles/hour

Then,

a=yb/(x-y)

a=4×22/(6-4)

a=88/2

a=44miles/hour

Then, the house distance from the mountain is

Distance=ax

Distance =44×6

Distance =264miles

4 0
4 years ago
Read 2 more answers
What average force is needed to accelerate a 7.00-gram pellet from rest to 155 m/s over a distance of 0.600 m along the barrel o
leonid [27]

Answer: The force needed is 140.22 Newtons.

Explanation:

The key assumption in this problem is that the acceleration is constant along the path of the barrel bringing the pellet from velocity 0 to 155 m/s. This means the velocity is linearly increasing in time.

The force exerted on the pellet is

F = m a

In order to calculate the acceleration, given the displacement d,  

d = \frac{1}{2}at^2\implies a=\frac{2d}{t^2}

we will need to determine the time t it took for the pellet to make the distance through the barrel of 0.6m. That time can be determined using the average velocity of the pellet while traveling through the barrel. Since the velocity is a linear function of time, as mentioned above, the average is easy to calculate as:

\overline{v}=\frac{1}{2}(v_{end}-v_{start})=\frac{1}{2}(155-0)\frac{m}{s}=77.5\frac{m}{s}

This value can be used to determine the time for the pellet through the barrel:

t = \frac{d}{\overline{v}}=\frac{0.6m}{77.5\frac{m}{s}}\approx0.00774s

Finally, we can use the above to calculate the force:

F = ma = m\frac{2d}{t^2} = 0.007kg\cdot \frac{2\cdot 0.6 m}{0.00774^2 s^2}\approx 140.22N



8 0
3 years ago
Describe how light is used in modern forms of communication.
Kobotan [32]

Answer:Visible light communication (VLC) is a wireless method that uses light emitted by LEDs to deliver networked, mobile, high-speed communication similar to Wi-Fi, leading to the term Li-Fi. It can be used as standalone solution or in a supplementary role to radio-frequency (RF) or cellular network communication

Explanation:

8 0
3 years ago
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