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kirza4 [7]
2 years ago
15

The voltage between two parallel plates separated by a distance of 3.0 cm is 120 v. the electric field between the plates is:___

______
Physics
1 answer:
soldier1979 [14.2K]2 years ago
4 0

The electric field between two plates is 40 V/m.

To find the electric field between two plates, the values are given are,

distance =  3 cm

Voltage v = 120 v

<h3>What is electric field between two plates?</h3>

           The electric field between plates is the area where the charges between two plates, influences can be seen. For example, if a charged particle is placed near any charged plate, the plate exerts an electric force to attract or repel the charged particle.

Formula is given as,

                        E = V / d (v/m)

Substituting the given values,

                           = 120 /3

                           =40 v/m

Hence, the answer is 40 v/m.

Learn more about electric field,

brainly.com/question/8971780

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Compared with cool air, warm air is:________
photoshop1234 [79]

Answer:

option (b)

Explanation:

The density of cool air is more so always falls downwards.

Compared to the cool air, the density of air is less and it tends to rise.

Thus, option (b) is correct.

3 0
3 years ago
Evolutionary psychology is a relatively new approach to psychology that has been especially influenced by
Andrews [41]

the answer is David Buss

5 0
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A cube has a density of 1900 kg/m 3 kg/m3 while at rest in the laboratory. What is the cube's density as measured by an experime
Deffense [45]

Answer:

4847.94844926 kg/m³

Explanation:

\rho' = Actual density of cube = 1900 kg/m³

\rho = Density change due to motion

v = Velocity of cube = 0.92c

c = Speed of light = 3\times 10^8\ m/s

Relativistic density is given by

\rho=\dfrac{\rho'}{\sqrt{1-\dfrac{v^2}{c^2}}}\\\Rightarrow \rho=\dfrac{1900}{\sqrt{1-\dfrac{0.92^2c^2}{c^2}}}\\\Rightarrow \rho=\dfrac{1900}{\sqrt{1-0.92^2}}\\\Rightarrow \rho=4847.94844926\ kg/m^3

The cube's density as measured by an experimenter in the laboratory is 4847.94844926 kg/m³

3 0
3 years ago
A 210Ω resistor is connected in a circuit with a 110V battery. What total amount of charge passes through a point in the circuit
irina1246 [14]

Answer: 62.86 coulombs

Explanation:

Resistance (R) = 210Ω

Voltage of battery (V) = 110V

total amount of charge (Q) = ?

Time (T) = 2 minutes

The SI unit of time is seconds so convert 2 minutes to seconds

(If 1 minute = 60 seconds

2 minutes = 2 x 60 = 120 seconds)

To get the total charge, first get the current (I) flowing in the circuit by applying the formula V = IR

110V = I x 210Ω

I = 110V/210Ω

I = 0.524 Amps

Then, apply the formula

Charge = current x time

i.e Q = IT

Q = 0.524 Amps x 120 seconds

Q = 62.86 coulombs

Thus, 62.86 coulombs of charge passes through the circuit.

6 0
4 years ago
What is the net charge of both pantothenate and phosphopantothenate in aqueous solution at ph 7?.
wlad13 [49]

The net charge of pantothenate is -1 whereas the net charge of phosphopantothenate is −2 in aqueous solution at pH 7.

<h3>What is the net charge of both pantothenate and phosphopantothenate?</h3>

At pH 7, the carboxylic acid of pantothenate will lose one hydrogen ion which leads to a net charge of −1 on pantothenate while on the other hand, the phosphate group of phosphopantothenate will lose two hydrogen ions (−2) so −2 charge appear on the phosphopantothenate.

So we can conclude that The net charge of pantothenate is -1 whereas the net charge of phosphopantothenate is −2 in aqueous solution at pH 7.

Learn more about charge here: brainly.com/question/25923373

#SPJ1

4 0
2 years ago
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