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3 years ago

6

A graph of gas pressure versus the number of particles in a container is straight line.which other relationship will have a simi

lar graph

1 answer:

natali 33 [55]3 years ago

8 0

The answer here could be pressure vs. temperature.

Think of PV=nRT. T is related to P the same way as n is (assuming all other variables are kept constant).

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What is the relationship between thickness of lens and focal length?

Ymorist [56]

Thick lens will have shorter and consequently thin lens will have greater focal length. Because, For a thick lens, the optical path length of the light is more, than for a thin lens, thus, the bending of light will be more in case of a thicker lens. Consequently, it has a shorter focal length.

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3 years ago

How many kisses is 50 hectokisses?<br><br> What would you be if you earn a megabuck year?

Keith_Richards [23]

Answer:

1.) 6 hectokisses 50 times

2.) Millionaire

Explanation:

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3 years ago

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You use 8x binoculars were used on a warbler (14cm long) in a tree 18cm away. What angle (in degrees) does the image of the warb

mafiozo [28]

Answer:

The angle it subtend on the retina is $\theta_z = 0.44586^o$

Explanation:

From the question we are told that

The length of the warbler is $L = 14cm = \frac{14}{100} = 0.14m$

The distance from the binoculars is $d = 18cm = \frac{18}{100} = 0.18m$

The magnification of the binoculars is $M =8$

Without the 8 X binoculars the angle made with the angular size of the object is mathematically represented as

$\theta = \frac{L}{d}$

$\theta = \frac{0.14}{0.18}$

$= 0.007778 rad$

Now magnification can be represented mathematically as

$M = \frac{\theta _z}{\theta}$

Where $\theta_z$ is the angle the image of the warbler subtend on your retina when the binoculars i.e the binoculars zoom.

So

$\theta_z = M * \theta$

=> $\theta_z =8 * 0.007778$

$= 0.0622222224$

Generally the conversion to degrees can be mathematically evaluated as

$\theta_z = 0.062222224 * (\frac{360 }{2 \pi rad} )$

$\theta_z = 0.44586^o$

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3 years ago

Calculate the standard electrode potential difference (e°) of the daniell cell (at 1 bar) if temperature is 473.15 k.

anzhelika [568]

Missing data in the text of the exercise: The molar concentration of Zinc is 10 times the molar concentration of copper.

Solution:

1) First of all, let's calculate the standard electrode potential difference at standard temperature. This is given by:
$E^0=E_{cat}^0-E_{an}^0$
where $E_{cat}^0$ is the standard potential at the cathode, while $E_{an}^0$ is the standard potential at the anode. For a Daniel Cell, at the cathode we have copper: $E_{Cu}^0=+0.34 V$ , while at the anode we have zinc: $E_{Zn}^0=-0.76 V$ . Therefore, at standard temperature the electrode potential difference of the Daniel Cell is
$E^0=+0.34 V-(-0.76 V)=+1.1 V$

2) To calculate $E^0$ at any temperature T, we should use Nerst equation:
$E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}$
where
$R=8.31 J/(K mol)$
$T=473.15 K$ is the temperature in our problem
$z=2$ is the number of electrons transferred in the cell's reaction
$F=9.65\cdot 10^4 C/mol$ is the Faraday's constant
$[Zn]$ and $[Cu]$ are the molar concentrations of zinc and in copper, and in our problem we have $[Zn]=10[Cu]$ .
Using all these data inside the equation, and using $E^0=+1.1 V$ , in the end we find:
$E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}=+1.053 V$

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3 years ago

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marshall27 [118]

A.The swimmer pushes the water
C. the walls force against the ball

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3 years ago

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