Answer:
potential difference V= 300 volts
Explanation:
Given:
d= 2.0 cm = 0.02m
E = 15 kN/C = 15 × 10³ N/C
For a uniform field between two plates, the Electric Filed Intensity (E) is proportional to the potential difference (V) and inversely proportional to distance between the plates.
E= V/d
⇒ V= E×d = 15 × 10³ N/C × 0.02 m = 300 volts (∴1 Nm/C = 1 J/C= 1 volts)
The atoms of some materials have no free electrons in their outer orbits. These electrons are busy doing other jobs, like being shared in the orbits of two adjacent atoms. They are so closely held that it is very difficult to pull them away. Most compounds of carbon and hydrogen are like this.
<span>Plastics, whose molecules are made from long combinations of carbon and hydrogen atoms, have few or no free electrons. This means that plastics are poor conductors of electricity (and they are also poor conductors of heat). hope that helped.</span>
0.00032cm*4.02=1.2864 × 10^-3 in scientific notation.
Explanation:
If you like my answer than please mark me brainliest thanks
<span>3.36x10^5 Pascals
The ideal gas law is
PV=nRT
where
P = Pressure
V = Volume
n = number of moles of gas particles
R = Ideal gas constant
T = Absolute temperature
Since n and R will remain constant, let's divide both sides of the equation by T, getting
PV=nRT
PV/T=nR
Since the initial value of PV/T will be equal to the final value of PV/T let's set them equal to each other with the equation
P1V1/T1 = P2V2/T2
where
P1, V1, T1 = Initial pressure, volume, temperature
P2, V2, T2 = Final pressure, volume, temperature
Now convert the temperatures to absolute temperature by adding 273.15 to both of them.
T1 = 27 + 273.15 = 300.15
T2 = 157 + 273.15 = 430.15
Substitute the known values into the equation
1.5E5*0.75/300.15 = P2*0.48/430.15
And solve for P2
1.5E5*0.75/300.15 = P2*0.48/430.15
430.15 * 1.5E5*0.75/300.15 = P2*0.48
64522500*0.75/300.15 = P2*0.48
48391875/300.15 = P2*0.48
161225.6372 = P2*0.48
161225.6372/0.48 = P2
335886.7441 = P2
Rounding to 3 significant figures gives 3.36x10^5 Pascals.
(technically, I should round to 2 significant figures for the result of 3.4x10^5 Pascals, but given the precision of the volumes, I suspect that the extra 0 in the initial pressure was accidentally omitted. It should have been 1.50e5 instead of 1.5e5).</span>
