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3 years ago

Please help ASAP please

Physics

1 answer:

Alex_Xolod [135]3 years ago

6 0

Answer:

Explanation:

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I don't understand this question at all, can I please get some help?

beks73 [17]

V^2/R=180W
v=root 180R

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3 years ago

The magnitude of the electrostatic force between two identical ions that are separated by a distance of 5.0A is 3.7×10^-9N.a) wh

jeyben [28]

Explanation:

Given that,

Electrostatic force, $F=3.7\times 10^{-9}\ N$

Distance, $r=5\ A=5\times 10^{-10}\ m$

(a) $F=\dfrac{kq^2}{r^2}$ , q is the charge on the ion

$q=\sqrt{\dfrac{Fr^2}{k}}$

$q=\sqrt{\dfrac{3.7\times 10^{-9}\times (5\times 10^{-10})^2}{9\times 10^9}}$

$q=3.2\times 10^{-19}\ C$

(b) Let n is the number of electrons are missing from each ion. It can be calculated as :

$n=\dfrac{q}{e}$

$n=\dfrac{3.2\times 10^{-19}}{1.6\times 10^{-19}}$

n = 2

Hence, this is the required solution.

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3 years ago

What of the following is the mathematical form of boyles law?

viva [34]

Answer:

Explanation:

V=1/p

By means of cross multiplication so by that we will have pv=1 which also implies p1v1=p2v2 coz boyles law states that the volume of a given mass of gas is inversely proportional to pressure provided that the temperature in kelvin remains constant

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A converging lens can produce both real and virtual images depending on the position of the object. Explain when converging lens

inessss [21]

Answer: C

Explanation:

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3 years ago

Read 2 more answers

Las condiciones iniciales de un gas son 3000 cm3

slava [35]

Answer:

T'=92.70°C

Explanation:

To find the temperature of the gas you use the equation for ideal gases:

$PV=nRT$

V: volume = 3000cm^3 = 3L

P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm

n: number of moles

R: ideal gas constant = 0.082 atm.L/mol.K

T: temperature = 27°C = 300.15K

For the given values you firs calculate the number n of moles:

$n=\frac{PV}{RT}=\frac{(1520[0.001315atm])(3L)}{(0.082\frac{atm.L}{mol.K})(300.15K)}=0.200moles$

this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:

$T'=\frac{P'V'}{nR}=\frac{(3atm)(2L)}{(0.200\ moles)(0.082\frac{atm.L}{mol.K})}=365.85K=92.70\°C$

hence, T'=92.70°C

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3 years ago