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3 years ago

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The function s(t) represents the position of an object at time t moving along a line. Suppose s (2 )equals 146and s (6 )equals

254.Find the average velocity of the object over the interval of time [2 comma 6 ].

1 answer:

sineoko [7]3 years ago

5 0

Answer:

v(t) = 27 units

Explanation:

The function s(t) represents the position of an object at time t moving along a line such that,

$s(2)=146$

and

$s(6)=254$

We need to find the average velocity of the object over the interval of time [2,6]. The velocity of the object is equal to the total distance divided by time. It is given by :

$v(t)=\dfrac{s(6)-s(2)}{6-2}$

$v(t)=\dfrac{254-146}{6-2}$

v(t) = 27 units

So, the average velocity of the object is 27 units. Hence, this is the required solution.

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Two trains approach each other on parallel tracks. Each has a speed of 95 kph relative to the ground. If they are initially 8.5

aliya0001 [1]

Answer: 2.83 minutes

Explanation:

It is understood that trains are approaching. That is, they have speeds of equal magnitude but opposite. When train A travels x meters northbound, then train B travels the same distance southbound.

Therefore trains approach at a speed of:

$V = 95 +95 = 190\ km / h$

Then:

$V = \frac{x}{t}\\\\t = \frac{x}{V}$

Where x is the distance between the trains

$x = 8.5\ km$

So the time in which both trains meet is:

$t =\frac{8.5\ km}{180\ \frac{km}{h}}\\\\t=0.04722\ hours$

This is:

$0.04722*\frac{60\ minutes}{1\ hour} = 2.83\ minutes$

<em />

<em>How long will it be before they reach one another ?</em>

<h3>2.83 minutes</h3>

3 0

3 years ago

The n = 2 to n = 6 transition in the bohr hydrogen atom corresponds to the ________ of a photon with a wavelength of ________ nm

Colt1911 [192]

The energy levels of the hydrogen atom are quantized and their energy is given by the approximated formula
$E=- \frac{13.6}{n^2} [eV]$
where n is the number of the level.

In the transition from n=2 to n=6, the variation of energy is
$\Delta E=E(n=6)-E(n=2)=-13.6 ( \frac{1}{6^2}- \frac{1}{2^2} )[eV]=3.02 eV$
Since this variation is positive, it means that the system has gained energy, so it must have absorbed a photon.

The energy of photon absorbed is equal to this $\Delta E$ . Converting it into Joule,
$\Delta E=3.02 eV=4.84 \cdot 10^{-19}J$
The energy of the photon is
$E=hf$
where h is the Planck constant while f is its frequency. Writing $\Delta E=hf$ , we can write the frequency f of the photon:
$f= \frac{\Delta E}{h}= \frac{4.84 \cdot 10^{-19}J}{6.63 \cdot 10^{-34}m^2 kg/s}=7.29 \cdot 10^{14}Hz$

The photon travels at the speed of light, $c=3 \cdot 10^8 m/s$ , so its wavelength is
$\lambda = \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{7.29 \cdot 10^{14}Hz}=4.11 \cdot 10^{-7}m=411 nm$

So, the initial sentence can be completed as:
The n = 2 to n = 6 transition in the bohr hydrogen atom corresponds to the "absorption" of a photon with a wavelength of "411" nm.

4 0

4 years ago

Read 2 more answers

Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi

Orlov [11]

The number converted is $0.0467 \frac{(kg)(s)}{m^3}$

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

$1 kg = 1000 g = 10^6 mg$

$1 min = 60 s$

$1 m^3 = 1000 L$

The original unit that we have is

$\frac{mg\cdot min}{L}$

Therefore, it can be rewritten as:

$=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot 60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}$

Therefore, since the initial number was 0.779, the final value is

$0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}$

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4 years ago

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tatuchka [14]

Explanation:

C,

.hahxxjdndjdndjgfndkndidjdodnxondos

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3 years ago

I'll Vote Brainliest

aliina [53]

B) only in animal cells

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