All categories

3 years ago

Compare and contrast speed and velocity

1 answer:

8 0

Where speed is distance/time, velocity is displacement/time.

What this means is that velocity is the length covered in relation to the starting point.
Speed is just the distance travelled no matter where you began.

When going around a circular track, you might have a speed value. However, since you get back to the same location at every lap, you have 0 velocity.

Hope I helped :)

You might be interested in

An escalator carries you from one level to the next in an airport terminal. The upper level is 4.4m above the lower level, and t

horrorfan [7]

Answer:

Part a)

$W = 2716.5 J$

Part B)

$W = -2716.5 J$

Explanation:

Part A)

While escalator is moving up work done to move the person upwards is given as

$W = mgh$

here we know that

m = 63 kg

h = 4.4 m

now we have

$W = 63 \times 9.8 \times 4.4$

$W = 2716.5 J$

Part B)

Work done while we move from up to down

So we have

$W = - mgh$

so we have

$W = -2716.5 J$

6 0

3 years ago

Compare and contrast infrasonic and ultrasonic vibrations

Olegator [25]

Answer:

Explanation:

Ultrasonic waves are acoustic waves that are so high in frequency that humans can't hear them; however, infrasonic waves are sound waves that are lower in frequency than what humans can hear. A subsonic wave is a wave that is traveling slower than the speed of sound and a supersonic waves travels faster

8 0

3 years ago

Water waves in a small tank are .06 m long. They pass a given point at a rate of 14.8 waves every three seconds. What is the spe

snow_lady [41]

Answer:

Speed = 0.296m/2

Period = 0.203 s

Explanation:

If by 'long' you mean the wavelength of the waves, then the wavelength $\lambda=0.06m$ .

The frequency $f$ of the waves is 14.8 waves every 3 seconds or

$f=14.8/3 =4.33Hz$ .

Now the relationship between wavelength $\lambda$ , frequency $f$ and speed $v$ of the waves is:

$v=\lambda f$

We put in the values $\lambda=0.06m$ and $f=4.933Hz$ and get:

$\boxed{v=0.06*4.922=0.296m/s}$

Now the period $T$ is just the inverse of the frequency, or

$T=\frac{1}{f}$

$\boxed{T=\frac{1}{4.933}=0.203\:seconds }$

4 0

3 years ago

A drag racer starts from rest and accelerates at 7.4 m/s2. How far will he travel in 2.0 seconds?

Leto [7]

Using the kinematic equation below we can determine the distance traveled if t=2, a=7.4m/s^2. First we must determine the final velocity:

$v_{final}=v_{initial}+\frac{1}{2}at\\\\v_{final}=0+(7.4m/s^2)(2s)=34.8m/s$

Now we will determine the distance traveled:

$v_{final}^2=v_{initial}^2+2a \Delta x\\\\\Delta x = \frac{v_{final}^2}{2a} =\frac{(34.8)^2}{(2)(7.4)}=81.83 m$

Therefore, the drag racer traveled 81.83 meters in 2 seconds.

3 0

3 years ago

What is the longest wavelength of radiation with enough energy to break carbon-carbon bonds?

iogann1982 [59]

The longest wavelength of radiation used to break carbon-carbon bonds is 344 nm.

<u>Explanation:</u>

The longest wavelength of radiation can also be stated as the minimum radiation frequency required to cut carbon-carbon bond should be equal to the threshold energy of the carbon-carbon bonds.

The threshold energy will be equal to the binding energy of the carbon-carbon bonds. As it is known that carbon-carbon bonds exhibit a binding energy of 348 kJ/mole, the threshold energy to break it, is determined as followed.

First, we have to convert the energy from kJ/mol to J, i.e., energy for the carbon-carbon molecules,

$\text { Energy } = \frac{348 \mathrm{KJ} / \mathrm{mol}}{6.023 \times 10^{23} \text { photons }} \times 1 \text { mole } \times 1000 = 57.77 \times 10^{-20} = 5.78 \times 10^{-19} J$

As,

$E=h v=\frac{h c}{\lambda}$

So,

$\lambda=\frac{h c}{E}=\frac{6.626 \times 10^{-34} \times 3 * 10^{8}}{5.78 \times 10^{-19}}=3.44 \times 10^{-7}$

Thus, $\lambda=344 \mathrm{nm}$ is the longest wavelength of radiation used to break carbon-carbon bonds.

5 0

3 years ago